BUG-FREE-For Dream
一直直到bug-free。不能错任何一点。
思路不清晰：刷两天。
做错了，刷一天。
直到bug-free。高亮，标红。
185，OA(YAMAXUN)---
(1) findFirstDuplicate string in a list of string.
import java.util.HashSet;
import java.util.Set;
public class Solution {
public static void main(String[] args) {
String[] strs = {"abc", "bd", "bd", "abc"};
System.out.println(findFirstDuplicate(strs));
}
public static String findFirstDuplicate(String[] strs) {
if (strs == null || strs.length == 0) {
return null;
}
Set<String> set = new HashSet();
for (int i = 0; i < strs.length; i++) {
if (set.contains(strs[i])) {
return strs[i];
} else {
set.add(strs[i]);
}
}
return null;
}
}
(2)merge two sorted array---
Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.
答案和思路：
public class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int i = m - 1;
int j = n - 1;
int index = m + n - 1;
while (i >= 0 && j >= 0) {
if (nums1[i] > nums2[j]) {
nums1[index--] = nums1[i--];
} else {
nums1[index--] = nums2[j--];
}
}
while (i >= 0) {
nums1[index--] = nums1[i--];
}
while (j >= 0) {
nums1[index--] = nums2[j--];
}
}
}
如果要求新开一个数组。
import java.util.Arrays;
public class Solution {
public static void main(String[] args) {
int[] a = {1, 3, 7};
int[] b = {2, 4, 5, 6};
System.out.println(Arrays.toString(merge(a, 3, b, 4)));
}
public static int[] merge(int[] nums1, int m, int[] nums2, int n) {
int index = 0;
int i = 0;
int j = 0;
int[] result = new int[m + n];
while (i < m && j < n) {
if (nums1[i] < nums2[j]) {
result[index++] = nums1[i++];
} else {
result[index++] = nums2[j++];
}
}
while (i < m) {
result[index++] = nums1[i++];
}
while (j < n) {
result[index++] = nums2[j++];
}
return result;
}
}
(3)algorithm crush
You are given a list of size , initialized with zeroes. You have to perform operations on the list and output the maximum of final values of all the elements in the list. For every operation, you are given three integers , and and you have to add value to all the elements ranging from index to (both inclusive). Input Format First line will contain two integers and separated by a single space. Next lines will contain three integers , and separated by a single space. Numbers in list are numbered from to . Constraints Output Format A single line containing maximum value in the updated list. Sample Input 5 3 1 2 100 2 5 100 3 4 100 Sample Output 200 Explanation After first update list will be 100 100 0 0 0. After second update list will be 100 200 100 100 100. After third update list will be 100 200 200 200 100. So the required answer will be 200.
我的答案：
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner in = new Scanner(System.in);
int m = in.nextInt();
int n = in.nextInt();
int[] a = new int[m];
for (int i = 0; i < n; i++) {
int left = in.nextInt();
int right = in.nextInt();
int value = in.nextInt();
for (int j = left - 1; j < right; j++) {
a[j] += value;
}
}
int max = Integer.MIN_VALUE;
for (int i = 0; i < m; i++) {
max = Math.max(max, a[i]);
}
System.out.println(max);
}
}
正确答案：
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner in = new Scanner(System.in);
int m = in.nextInt();
int n = in.nextInt();
int[] a = new int[m];
for (int i = 0; i < n; i++) {
int left = in.nextInt();
int right = in.nextInt();
int value = in.nextInt();
a[left - 1] += value;
if (right < m - 1) {
a[right] -= value;
}
}
int max = Integer.MIN_VALUE;
int sum = 0;
for (int i = 0; i < m; i++) {
sum += a[i];
a[i] = sum;
max = Math.max(a[i], max);
}
System.out.println(max);
}
}
(4) valid parentheses---
Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine if the input string is valid.
The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.
Subscribe to see which companies asked this question
答案：
public class Solution {
public boolean isValid(String s) {
if (s == null || s.length() == 0) {
return true;
}
Stack<Character> stack = new Stack();
for (int i = 0; i < s.length(); i++) {
char cur = s.charAt(i);
if (cur == '(' || cur == '{' || cur == '[') {
stack.push(s.charAt(i));
} else {
if (stack.isEmpty()) {
return false;
}
if (cur == ')' && stack.pop() != '(') {
return false;
} else if (cur == ']' && stack.pop() != '[') {
return false;
} else if (cur == '}' && stack.pop() != '{') {
return false;
}
}
}
if (stack.isEmpty()) {
return true;
}
return false;
}
}
（5）lowest common ancestor
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according
答案和思路：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || p == root || q == root) {
return root;
}
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left != null && right != null) {
return root;
} else if (left != null) {
return left;
} else {
return right;
}
}
}
（6）find first repeated word in a string
public static Character getFirstNotRepeatedChar(String input) {
byte[] flags = new byte[256]; //all is initialized by 0
for (int i = 0; i < input.length(); i++) { // O(n)
flags[(int)input.charAt(i)]++ ;
}
for (int i = 0; i < input.length(); i++) { // O(n)
if(flags[(int)input.charAt(i)] == 1)
return input.charAt(i);
}
return null;
}
如果是找word可以split多次。
import java.util.HashSet;
import java.util.Scanner;
public class Solution {
public static void main(String[] args) {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
Scanner in = new Scanner(System.in);
String str = in.nextLine();
String[] words = str.split("\t|-|\\.|:");
// String[] words = str.split("\\.");
HashSet<String> set = new HashSet();
for (int i = 0; i < words.length; i++) {
if (set.contains(words[i])) {
System.out.println(words[i]);
} else {
set.add(words[i]);
}
}
}
}
（7）找公共祖先
1，BST tree。
public class Solution {
public boolean isValid(String s) {
if (s == null || s.length() == 0) {
return true;
}
Stack<Character> stack = new Stack();
for (int i = 0; i < s.length(); i++) {
char cur = s.charAt(i);
if (cur == '(' || cur == '{' || cur == '[') {
stack.push(s.charAt(i));
} else {
if (stack.isEmpty()) {
return false;
}
if (cur == ')' && stack.pop() != '(') {
return false;
} else if (cur == ']' && stack.pop() != '[') {
return false;
} else if (cur == '}' && stack.pop() != '{') {
return false;
}
}
}
if (stack.isEmpty()) {
return true;
}
return false;
}
}
2，binary tree
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || root == p || root == q) {
return root;
}
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left == null) {
return right;
} else if (right == null) {
return left;
} else {
return root;
}
}
}
3，parent binary tree
/**
* Definition of ParentTreeNode:
*
* class ParentTreeNode {
* public ParentTreeNode parent, left, right;
* }
*/
public class Solution {
/**
* @param root: The root of the tree
* @param A, B: Two node in the tree
* @return: The lowest common ancestor of A and B
*/
public ParentTreeNode lowestCommonAncestorII(ParentTreeNode root,
ParentTreeNode A,
ParentTreeNode B) {
// Write your code here
List<ParentTreeNode> pathA = getPathToRoot(A);
List<ParentTreeNode> pathB = getPathToRoot(B);
int indexA = pathA.size() - 1;
int indexB = pathB.size() - 1;
ParentTreeNode lowestAncestor = null;
while (indexA >= 0 && indexB >= 0) {
if (pathA.get(indexA) != pathB.get(indexB)) {
break;
}
lowestAncestor = pathA.get(indexA);
indexA--;
indexB--;
}
return lowestAncestor;
}
private static List<ParentTreeNode> getPathToRoot(ParentTreeNode node) {
List<ParentTreeNode> path = new ArrayList();
while (node != null) {
path.add(node);
node = node.parent;
}
return path;
}
}
4,multi branch tree
import java.util.ArrayList;
import java.util.List;
class TreeNode {
int val;
List<TreeNode> children = new ArrayList();
public TreeNode(int v) {
val = v;
}
}
public class Solution {
public static void main(String[] args) {
TreeNode root = new TreeNode(1);
root.children.add(new TreeNode(2));
root.children.add(new TreeNode(3));
root.children.get(0).children.add(new TreeNode(4));
System.out.println(getLowestCommonAncestor(root, root, root).val);
System.out.println(getLowestCommonAncestor(root, root.children.get(0), root.children.get(0).children.get(0)).val);
}
public static TreeNode getLowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) {
return null;
}
List<TreeNode> pPath = new ArrayList();
List<TreeNode> qPath = new ArrayList();
getPath(root, p, pPath);
getPath(root, q, qPath);
//
if (pPath.isEmpty() && qPath.isEmpty()) {
return null;
}
int index = 0;
TreeNode result = null;
while (index < pPath.size() && index < qPath.size()) {
if (pPath.get(index) == qPath.get(index)) {
result = pPath.get(index);
index++;
} else {
break;
}
}
return result;
}
public static boolean getPath(TreeNode root, TreeNode target, List<TreeNode> path) {
path.add(root);
if (root == target) {
return true;
}
boolean found = false;
List<TreeNode> list = root.children;
for (TreeNode n : list) {
if (found) {
break;
}
found = getPath(n, target, path);
}
if (!found) {
path.remove(path.size() - 1);
}
return found;
}
}
employee manager:
import java.util.ArrayList;
import java.util.List;
class TreeNode {
int val;
List<TreeNode> children = new ArrayList();
public TreeNode(int v) {
val = v;
}
}
public class Solution {
public static void main(String[] args) {
TreeNode root = new TreeNode(1);
root.children.add(new TreeNode(2));
root.children.add(new TreeNode(3));
root.children.get(0).children.add(new TreeNode(4));
System.out.println(getLowestCommonAncestor(root, root, root).val);
System.out.println(getLowestCommonAncestor(root, root.children.get(0), root.children.get(0).children.get(0)).val);
}
public static TreeNode getLowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) {
return null;
}
List<TreeNode> pPath = new ArrayList();
List<TreeNode> qPath = new ArrayList();
boolean pFlag = getPath(root, p, pPath);
boolean qFlag = getPath(root, q, qPath);
//
if (pPath.isEmpty() && qPath.isEmpty()) {
return null;
}
int index = 0;
TreeNode result = null;
while (index < pPath.size() && index < qPath.size()) {
if (pPath.get(index) == qPath.get(index)) {
result = pPath.get(index);
index++;
} else {
break;
}
}
return result;
}
public static boolean getPath(TreeNode root, TreeNode target, List<TreeNode> path) {
path.add(root);
if (root == null) {
return false;
}
if (root == target) {
return true;
}
boolean found = false;
List<TreeNode> list = root.children;
for (TreeNode n : list) {
if (found) {
break;
}
found = getPath(n, target, path);
}
if (!found) {
path.remove(path.size() - 1);
}
return found;
}
}
注意如果是root就返回null。要看题目的要求。
import java.util.ArrayList;
import java.util.List;
class TreeNode {
int val;
List<TreeNode> children = new ArrayList();
public TreeNode(int v) {
val = v;
}
}
public class Solution {
public static void main(String[] args) {
TreeNode root = new TreeNode(1);
root.children.add(new TreeNode(2));
root.children.add(new TreeNode(3));
root.children.get(0).children.add(new TreeNode(4));
System.out.println(getLowestCommonAncestor(root, root, root));
System.out.println(getLowestCommonAncestor(root, root.children.get(0), root.children.get(0).children.get(0)).val);
}
public static TreeNode getLowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) {
return null;
}
List<TreeNode> pPath = new ArrayList();
List<TreeNode> qPath = new ArrayList();
boolean pFlag = getPath(root, p, pPath);
boolean qFlag = getPath(root, q, qPath);
//
if (!pFlag || !qFlag) {
return null;
}
if (pPath.isEmpty() && qPath.isEmpty() || pPath.size() == 1 || qPath.size() == 1) {
return null;
}
int index = 0;
TreeNode result = null;
while (index < pPath.size() && index < qPath.size()) {
if (pPath.get(index) == qPath.get(index)) {
result = pPath.get(index);
index++;
} else {
break;
}
}
return result;
}
public static boolean getPath(TreeNode root, TreeNode target, List<TreeNode> path) {
path.add(root);
if (root == null) {
return false;
}
if (root == target) {
return true;
}
boolean found = false;
List<TreeNode> list = root.children;
for (TreeNode n : list) {
if (found) {
break;
}
found = getPath(n, target, path);
}
if (!found) {
path.remove(path.size() - 1);
}
return found;
}
}
184,backpack---http://www.lintcode.com/zh-cn/problem/backpack/---NOT BUG FREE
Given n items with size Ai, an integer m denotes the size of a backpack. How full you can fill this backpack?
答案和思路：dp[i][j]表示前i个物品是否能够构成
public class Solution {
/**
* @param m: An integer m denotes the size of a backpack
* @param A: Given n items with size A[i]
* @return: The maximum size
*/
public int backPack(int m, int[] A) {
// write your code here
if (A == null || A.length == 0) {
return 0;
}
// dp[i][j] : 前i个物品是否能够组成j价值
boolean[][] dp = new boolean[2][m + 1];
dp[0][0] = true;
for (int i = 0; i < A.length; i++) {
for (int j = 0; j < m + 1; j++) {
dp[(i + 1) % 2][j] = dp[i % 2][j];
if (j >= A[i] && dp[i % 2][j - A[i]]) {
dp[(i + 1) % 2][j] = true;
}
}
}
for (int j = m; j >= 0; j--) {
if (dp[(A.length) % 2][j]) {
return j;
}
}
return 0;
}
}
183, maximum product subarray---https://leetcode.com/problems/maximum-product-subarray/---BUG FREE
Find the contiguous subarray within an array (containing at least one number) which has the largest product. For example, given the array [2,3,-2,4], the contiguous subarray [2,3] has the largest product = 6.
答案和思路：
public class Solution {
public int maxProduct(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int len = nums.length;
int[] max = new int[len];
int[] min = new int[len];
max[0] = min[0] = nums[0];
int result = nums[0];
for (int i = 1; i < nums.length; i++) {
min[i] = Math.min(nums[i], Math.min(min[i - 1] * nums[i], max[i - 1] * nums[i]));
max[i] = Math.max(nums[i], Math.max(min[i - 1] * nums[i], max[i - 1] * nums[i]));
result = Math.max(result, max[i]);
}
return result;
}
}
182，maximal square---http://www.lintcode.com/en/problem/maximal-square/--BUG FREE
Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.
答案和思路：以这个点为结尾的正方形取决于他旁边的三个点中最小的那个+1.
public class Solution {
/**
* @param matrix: a matrix of 0 and 1
* @return: an integer
*/
public int maxSquare(int[][] matrix) {
// write your code here
if (null == matrix || matrix.length == 0) {
return 0;
}
int m = matrix.length;
int n = matrix[0].length;
int max = 0;
int[][] dp = new int[2][n]; // dp[i][j]表示以i，j结尾能达到的最大的正方形的边长
for (int i = 0; i < m; ++i) {
if (matrix[i][0] == 1) {
max = 1;
}
dp[i % 2][0] = matrix[i][0];
}
for (int j = 0; j < n; ++j) {
if (dp[0][j] == 1) {
max = 1;
}
dp[0][j] = matrix[0][j];
}
for (int i = 1; i < m; ++i) {
for (int j = 1; j < n; ++j) {
if (matrix[i][j] == 1) {
dp[i % 2][j] = Math.min(dp[(i - 1) % 2][j - 1], Math.min(dp[(i - 1) % 2][j], dp[i % 2][j - 1])) + 1;.
} else {
dp[i % 2][j] = 0;
}
max = Math.max(max, dp[i % 2][j]);
}
}
return max * max;
}
}
181，house robber II---https://leetcode.com/problems/house-robber-ii/---BUG FREE
Note: This is an extension of House Robber. After robbing those houses on that street, the thief has found himself a new place for his thievery so that he will not get too much attention. This time, all houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, the security system for these houses remain the same as for those in the previous street. Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the poli
答案和思路：
public class Solution {
public int rob(int[] nums) {
if (nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
return Math.max(robber1(nums, 0, nums.length - 2), robber1(nums, 1, nums.length - 1));
}
public int robber1(int[] nums, int st, int ed) {
int []res = new int[2];
if(st == ed)
return nums[ed];
if(st+1 == ed)
return Math.max(nums[st], nums[ed]);
res[st%2] = nums[st];
res[(st+1)%2] = Math.max(nums[st], nums[st+1]);
for(int i = st+2; i <= ed; i++) {
res[i%2] = Math.max(res[(i-1)%2], res[(i-2)%2] + nums[i]);
}
return res[ed%2];
}
}
180， house robber---https://leetcode.com/problems/house-robber/---bug FREE
You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security system connected and it will automatically contact the police if two adjacent houses were broken into on the same night. Given a list of non-negative integers representing the amount of money of each house, determine the maximum amount of money you can rob tonight without alerting the police.
答案和思路：
public class Solution {
public int rob(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length <= 2) {
return Math.max(nums[0], nums[nums.length - 1]);
}
int[] dp = new int[2];
dp[0] = nums[0];
dp[1] = Math.max(nums[0], nums[1]);
for (int i = 2; i < nums.length; i++) {
dp[i % 2] = Math.max(dp[(i - 1) % 2], nums[i] + dp[(i - 2) % 2]);
}
return dp[(nums.length - 1) % 2];
}
}
179，topological sorting---http://www.lintcode.com/zh-cn/problem/topological-sorting/---NOT BUG FREE
Given an directed graph, a topological order of the graph nodes is defined as follow: For each directed edge A -> B in graph, A must before B in the order list. The first node in the order can be any node in the graph with no nodes direct to it. Find any topological order for the given graph. Notice You can assume that there is at least one topological order in the graph. Have you met this question in a real interview? Yes Clarification Learn more about representation of graphs Example For graph as follow: picture The topological order can be: [0, 1, 2, 3, 4, 5] [0, 2, 3, 1, 5, 4] ...
答案和思路：
/**
* Definition for Directed graph.
* class DirectedGraphNode {
* int label;
* ArrayList<DirectedGraphNode> neighbors;
* DirectedGraphNode(int x) { label = x; neighbors = new ArrayList<DirectedGraphNode>(); }
* };
*/
public class Solution {
/**
* @param graph: A list of Directed graph node
* @return: Any topological order for the given graph.
*/
public ArrayList<DirectedGraphNode> topSort(ArrayList<DirectedGraphNode> graph) {
ArrayList<DirectedGraphNode> result = new ArrayList();
if (graph == null || graph.size() == 0) {
return result;
}
HashMap<DirectedGraphNode, Integer> map = new HashMap();
//map 用来记录那些成为了别人邻居的点作为多少个邻居，也就是他的入度是多少。
for (DirectedGraphNode g : graph) {
for (DirectedGraphNode n : g.neighbors) {
if (map.containsKey(n)) {
map.put(n, map.get(n) + 1);
} else {
map.put(n, 1);
}
}
}
Queue<DirectedGraphNode> queue = new LinkedList();
for (DirectedGraphNode g : graph) {
if (!map.containsKey(g)) {
//不在map里面，说明他的度为0
result.add(g);
queue.add(g);
}
}
while (!queue.isEmpty()) {
DirectedGraphNode node = queue.poll();
for (DirectedGraphNode n : node.neighbors) {
map.put(n, map.get(n) - 1);
if (map.get(n) == 0) {
result.add(n);
queue.add(n);
}
}
}
return result;
}
}
178,ugly number---https://leetcode.com/problems/ugly-number/---NOT BUG FREE
Write a program to check whether a given number is an ugly number. Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7. Note that 1 is typically treated as an ugly number.
答案和思路：
public class Solution {
/**
* @param num an integer
* @return true if num is an ugly number or false
*/
public boolean isUgly(int num) {
if (num <= 0) return false;
if (num == 1) return true;
while (num >= 2 && num % 2 == 0) num /= 2;
while (num >= 3 && num % 3 == 0) num /= 3;
while (num >= 5 && num % 5 == 0) num /= 5;
return num == 1;
}
}
177, median of three array---NOT BUG FREE
import java.util.ArrayList;
import java.util.Collections;
public class Solution {
/**
* @author ChunyueLi--Big Data Scientist
* @答案和思路：这道题也是希望利用二分思想来做，每一次找到排除掉的那一半。
* @复杂度：log(len)
*/
public static void main(String[] args) {
// int[] nums1 = {1, 2, 3, 4, 4};
// int[] nums2 = {0, 0, 4, 4, 4};
// int[] nums3 = {0, 0, 4, 4, 4};
int[] nums1 = {1, 12, 13};
int[] nums2 = {6, 7};
int[] nums3 = {0, 3, 4};
System.out.println(findMedianSortedArrays(nums1, nums2, nums3));
}
public static double findMedianSortedArrays(int[] nums1, int[] nums2, int[] nums3) {
if (nums1 == null || nums2 == null || nums3 == null) {
return -1;
}
int len1 = nums1.length;
int len2 = nums2.length;
int len3 = nums3.length;
int len = len1 + len2 + len3;
if (len % 2 == 1) {
return findKthFromThreeArray(nums1, 0, nums2, 0, nums3, 0, len / 2 + 1);
} else {
System.out.println("here");
System.out.println(findKthFromThreeArray(nums1, 0, nums2, 0, nums3, 0, len / 2));
System.out.println(findKthFromThreeArray(nums1, 0, nums2, 0, nums3, 0, len / 2 + 1));
return (findKthFromThreeArray(nums1, 0, nums2, 0, nums3, 0, len / 2) + findKthFromThreeArray(nums1, 0, nums2, 0, nums3, 0, len / 2 + 1)) / 2.0;
}
}
private static double findKthFromThreeArray(int[] a, int aIndex, int[] b, int bIndex, int[] c, int cIndex, int k) {
//如果其中一个走完了就退化成两个数组找
if (aIndex >= a.length) {
return findKthFromTwoArray(b, bIndex, c, cIndex, k);
}
if (bIndex >= b.length) {
return findKthFromTwoArray(a, aIndex, c, cIndex, k);
}
if (cIndex >= c.length) {
return findKthFromTwoArray(a, aIndex, b, bIndex, k);
}
if (k == 1) {
return Math.min(a[aIndex], Math.min(b[bIndex], c[cIndex]));
}
if (k == 2) { //这是最关键的地方，因为如果k == 2不进行判断，那么下一次移位就是0，会导致死循环，从而爆栈。
ArrayList<Integer> temp = new ArrayList();
for (int i = 0; i < 2; i++) {
if (aIndex < a.length) {
temp.add(a[aIndex++]);
}
if (bIndex < b.length) {
temp.add(b[bIndex++]);
}
if (cIndex < c.length) {
temp.add(c[cIndex++]);
}
}
Collections.sort(temp);
return temp.get(1);
}
int aMid = aIndex + k / 3 - 1 < a.length ? a[aIndex + k / 3 - 1] : Integer.MAX_VALUE;
int bMid = bIndex + k / 3 - 1 < b.length ? b[bIndex + k / 3 - 1] : Integer.MAX_VALUE;
int cMid = cIndex + k / 3 - 1 < c.length ? c[cIndex + k / 3 - 1] : Integer.MAX_VALUE;
if (aMid < bMid) {
if (aMid < cMid) {
return findKthFromThreeArray(a, aIndex + k / 3, b, bIndex, c, cIndex, k - k / 3);
} else {
return findKthFromThreeArray(a, aIndex, b, bIndex, c, cIndex + k / 3, k - k / 3);
}
} else {
if (bMid < cMid) {
return findKthFromThreeArray(a, aIndex, b, bIndex + k / 3, c, cIndex, k - k / 3);
} else {
return findKthFromThreeArray(a, aIndex, b, bIndex, c, cIndex + k / 3, k - k / 3);
}
}
}
private static double findKthFromTwoArray(int[] a, int aIndex, int[] b, int bIndex, int k) {
if (aIndex >= a.length) {
return b[bIndex + k - 1];
}
if (bIndex >= b.length) {
return a[aIndex + k - 1];
}
if (k == 1) {
return Math.min(a[aIndex], b[bIndex]);
}
int aMid = aIndex + k / 2 - 1 < a.length ? a[aIndex + k / 2 - 1] : Integer.MAX_VALUE;
int bMid = bIndex + k / 2 - 1 < b.length ? b[bIndex + k / 2 - 1] : Integer.MAX_VALUE;
if (aMid < bMid) {
return findKthFromTwoArray(a, aIndex + k / 2, b, bIndex, k - k / 2);
} else {
return findKthFromTwoArray(a, aIndex, b, bIndex + k / 2, k - k / 2);
}
}
}
178，palindrome partitioning---https://leetcode.com/problems/palindrome-partitioning/---NOT BUG FREE
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab", Return [ ["aa","b"], ["a","a","b"] ]
答案和思路：
public class Solution {
public List<List<String>> partition(String s) {
List<List<String>> results = new ArrayList();
if (s == null || s.length() == 0) {
return results;
}
ArrayList cur = new ArrayList();
partition(results, s, 0, cur);
return results;
}
private static void partition(List<List<String>> results, String s, int start, List<String> path) {
if (start == s.length()) {
results.add(new ArrayList(path));
return;
}
for (int i = start + 1; i <= s.length(); i++) {
String sub = s.substring(start, i);
if (isPalindrome(sub)) {
path.add(sub);
partition(results, s, i, path);
path.remove(path.size() - 1);
}
}
}
private static boolean isPalindrome(String s) {
for (int i = 0, j = s.length() - 1; i < j; i++, j--) {
if (s.charAt(i) != s.charAt(j)) {
return false;
}
}
return true;
}
}
177, longest consecutive sequence---https://leetcode.com/problems/longest-consecutive-sequence/---NOT BUG FREE
Given an unsorted array of integers, find the length of the longest consecutive elements sequence. For example, Given [100, 4, 200, 1, 3, 2], The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4. Your algorithm should run in O(n) complexity.
答案和思路：
public class Solution {
public int longestConsecutive(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
Set<Integer> set = new HashSet();
for (int i = 0; i < nums.length; i++) {
set.add(nums[i]);
}
int result = 0;
for (int i = 0; i < nums.length; i++) {
int left = nums[i] - 1;
while (set.contains(left)) {
set.remove(left);
left--;
}
int right = nums[i] + 1;
while (set.contains(right)) {
set.remove(right);
right++;
}
result = Math.max(result, right - left - 1);
}
return result;
}
}
176,search range in binary tree---http://www.lintcode.com/zh-cn/problem/search-range-in-binary-search-tree/#--NOT BUG FREE
Given two values k1 and k2 (where k1 < k2) and a root pointer to a Binary Search Tree. Find all the keys of tree in range k1 to k2. i.e. print all x such that k1<=x<=k2 and x is a key of given BST. Return all the keys in ascending order.
Have you met this question in a real interview? Yes
Example
If k1 = 10 and k2 = 22, then your function should return [12, 20, 22].
20
/ \
8 22
/ \
4 12
答案和思路：
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of the binary search tree.
* @param k1 and k2: range k1 to k2.
* @return: Return all keys that k1<=key<=k2 in ascending order.
*/
public ArrayList<Integer> searchRange(TreeNode root, int k1, int k2) {
// write your code here
ArrayList<Integer> result = new ArrayList();
if (root == null) {
return result;
}
searchRange(root, k1, k2, result);
return result;
}
private static void searchRange(TreeNode root, int k1, int k2, ArrayList<Integer> result) {
if (root == null) {
return;
}
if (root.val >= k1 && root.val <= k2) {
searchRange(root.left, k1, k2, result);
result.add(root.val);
searchRange(root.right, k1, k2, result);
} else if (root.val > k2) {
searchRange(root.left, k1, k2, result);
} else {
searchRange(root.right, k1, k2, result);
}
}
}
175, reverse string---https://leetcode.com/problems/reverse-string/---NOT BUG FREE
Write a function that takes a string as input and returns the string reversed. Example: Given s = "hello", return "olleh".
答案和思路：
public class Solution {
public String reverseString(String s) {
char[] c = s.toCharArray();
int left = 0;
int right = s.length() - 1;
while (left < right) {
char temp = c[left];
c[left] = c[right];
c[right] = temp;
left++;
right--;
}
return new String(c);
}
}
174，find minimum in rotated sorted array---https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/---NOT BUG FREE
Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array.
答案和思路：注意找到了就先跳出来，免得错过。
public class Solution {
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
if (nums[0] <= nums[nums.length - 1]) {
return nums[0];
}
int result = 0;
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (mid > 0 && nums[mid] < nums[mid - 1]) {
return nums[mid];
}
if (nums[mid] <= nums[nums.length - 1]) {
result = nums[mid];
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
}public class Solution {
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
if (nums[0] <= nums[nums.length - 1]) {
return nums[0];
}
int result = 0;
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (mid > 0 && nums[mid] < nums[mid - 1]) {
return nums[mid];
}
if (nums[mid] <= nums[nums.length - 1]) {
result = nums[mid];
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
}
173,find minimum in rotated sorted array-- https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/---NOT BUG FREE
Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array.
答案和思路：
public class Solution {
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
if (nums[0] <= nums[nums.length - 1]) {
return nums[0];
}
int result = 0;
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] <= nums[nums.length - 1]) {
result = nums[mid];
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
}
172，search a 2D matrix---https://leetcode.com/problems/search-a-2d-matrix-ii/---NOT BUG FREE
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted in ascending from left to right. Integers in each column are sorted in ascending from top to bottom. For example, Consider the following matrix: [ [1, 4, 7, 11, 15], [2, 5, 8, 12, 19], [3, 6, 9, 16, 22], [10, 13, 14, 17, 24], [18, 21, 23, 26, 30] ] Given target = 5, return true. Given target = 20, return false.
答案和思路：找到一个有序的。那么|__这个左下角的点就是有序的。
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) {
return false;
}
int x = matrix.length - 1;
int y = 0;
while (y < matrix[0].length && x >= 0) {
if (matrix[x][y] == target) {
return true;
} else if (matrix[x][y] > target) {
x--;
} else {
y++;
}
}
return false;
}
}
171,search a 2D matrix---https://leetcode.com/problems/search-a-2d-matrix/---NOT BUG FREE
Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: Integers in each row are sorted from left to right. The first integer of each row is greater than the last integer of the previous row. For example, Consider the following matrix: [ [1, 3, 5, 7], [10, 11, 16, 20], [23, 30, 34, 50] ] Given target = 3, return true.
答案和思路：
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) {
return false;
}
// find row
int row = 0;
int low = 0;
int high = matrix.length - 1;
while (low <= high) {
int mid = low + (high - low) / 2;
if (matrix[mid][0] == target) {
return true;
} else if (matrix[mid][0] < target) {
row = mid;
low = mid + 1;
} else {
high = mid - 1;
}
}
// find target
int left = 0;
int right = matrix[0].length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (matrix[row][mid] == target) {
return true;
} else if (matrix[row][mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return false;
}
}
170，sort list---
// version 1: Merge Sort
public class Solution {
private ListNode findMiddle(ListNode head) {
ListNode slow = head, fast = head.next;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
private ListNode merge(ListNode head1, ListNode head2) {
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
while (head1 != null && head2 != null) {
if (head1.val < head2.val) {
tail.next = head1;
head1 = head1.next;
} else {
tail.next = head2;
head2 = head2.next;
}
tail = tail.next;
}
if (head1 != null) {
tail.next = head1;
} else {
tail.next = head2;
}
return dummy.next;
}
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode mid = findMiddle(head);
ListNode right = sortList(mid.next);
mid.next = null;
ListNode left = sortList(head);
return merge(left, right);
}
}
// version 2: Quick Sort 1
public class Solution {
public ListNode sortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode mid = findMedian(head); // O(n)
ListNode leftDummy = new ListNode(0), leftTail = leftDummy;
ListNode rightDummy = new ListNode(0), rightTail = rightDummy;
ListNode middleDummy = new ListNode(0), middleTail = middleDummy;
while (head != null) {
if (head.val < mid.val) {
leftTail.next = head;
leftTail = head;
} else if (head.val > mid.val) {
rightTail.next = head;
rightTail = head;
} else {
middleTail.next = head;
middleTail = head;
}
head = head.next;
}
leftTail.next = null;
middleTail.next = null;
rightTail.next = null;
ListNode left = sortList(leftDummy.next);
ListNode right = sortList(rightDummy.next);
return concat(left, middleDummy.next, right);
}
private ListNode findMedian(ListNode head) {
ListNode slow = head, fast = head.next;
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
private ListNode concat(ListNode left, ListNode middle, ListNode right) {
ListNode dummy = new ListNode(0), tail = dummy;
tail.next = left; tail = getTail(tail);
tail.next = middle; tail = getTail(tail);
tail.next = right; tail = getTail(tail);
return dummy.next;
}
private ListNode getTail(ListNode head) {
if (head == null) {
return null;
}
while (head.next != null) {
head = head.next;
}
return head;
}
}
169，best time to buy and sell stock IV--- https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iv/---NOT BUG FREE
Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete at most k transactions.
答案和思路：
public class Solution{
//思路：
//localProfit[i][j]：到了第i天，并且第i天进行了第j次交易的maxProfit
//globalProfit[i][j]:到了第i天，进行了j次交易的maxProfit
//note：为什么要分两个？答：首先一定要有的是globalProfit[i][j]。因为这个就是要求的答案
//但是这个答案分为两种方案：1，是i天不做交易，那么他就==global[i-1][j];
//一种是，i天做交易。那么就要用一个i天做交易得到的maxProfit。
//然后最后max两种情况。
//【错误】注意，n天最多交易n/2次，当k很大，大于n/2要单独判断（相当于第一题，随便几次都可以），并且不要忘记那时候就要return
public int maxProfit(int k, int[] prices){
int n = prices.length;
if(k <= 0 || n == 0) return 0;
if(k >= n / 2){
int max = 0;
for(int i = 1; i < n; i++){
max += Math.max(0, prices[i] - prices[i - 1]);
}
return max;
}
int[][] localProfit = new int[n][k+1];
int[][] globalProfit = new int[n][k + 1];
//最重要的部分，从这里开始：
for(int j = 1; j <= k; ++j){
for(int i = 1; i < n; ++i){//天数是从0开始的。localProfit[0][0] = 0;
localProfit[i][j] = Math.max(localProfit[i - 1][j] + prices[i] - prices[i - 1],globalProfit[i - 1][j - 1] +Math.max(0, prices[i] - prices[i - 1]));
globalProfit[i][j] = Math.max(localProfit[i][j], globalProfit[i - 1][j]);
}
}
return globalProfit[n - 1][k];
}
}
168,best time to buy and sell stock III---https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/---NOT BUG FREE
Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete at most two transactions.
答案和思路：left，right两边来考虑。
public class Solution {
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) {
return 0;
}
int[] left = new int[prices.length];
int[] right = new int[prices.length];
int min = prices[0];
for (int i = 1; i < prices.length; i++) {
min = Math.min(prices[i], min);
left[i] = Math.max(left[i - 1], prices[i] - min);
}
int max = prices[prices.length - 1];
for (int i = prices.length - 2; i >= 0; i--) {
max = Math.max(prices[i], max);
right[i] = Math.max(right[i + 1], max - prices[i]);
}
int profit = 0;
for (int i = 0; i < prices.length; i++) {
profit = Math.max(left[i] + right[i], profit);
}
return profit;
}
}
167,best time to buy and sell stock II---https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/--BUG FREE
Say you have an array for which the ith element is the price of a given stock on day i. Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times). However, you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
答案和思路：
public class Solution {
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) {
return 0;
}
int result = 0;
for (int i = 1; i < prices.length; i++) {
result += Math.max(0, prices[i] - prices[i - 1]);
}
return result;
}
}
166，best time to buy and sell stock I ---https://leetcode.com/problems/best-time-to-buy-and-sell-stock/---BUG FREE
Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. Example 1: Input: [7, 1, 5, 3, 6, 4] Output: 5 max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price) Example 2: Input: [7, 6, 4, 3, 1] Output: 0 In this case, no transaction is done, i.e. max profit = 0.
答案和思路：贪心就行。
public class Solution {
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) {
return 0;
}
int min = prices[0];
int result = 0;
for (int i = 1; i < prices.length; i++) {
min = Math.min(min, prices[i]);
result = Math.max(result, prices[i] - min);
}
return result;
}
}
165,copy list with random pointer---https://leetcode.com/problems/copy-list-with-random-pointer/---NOT BUG FREE
A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null. Return a deep copy of the list.
答案和思路：
public class Solution {
private void copyNext(RandomListNode head) {
while (head != null) {
RandomListNode newNode = new RandomListNode(head.label);
newNode.random = head.random;
newNode.next = head.next;
head.next = newNode;
head = head.next.next;
}
}
private void copyRandom(RandomListNode head) {
while (head != null) {
if (head.next.random != null) {
head.next.random = head.random.next;
}
head = head.next.next;
}
}
private RandomListNode splitList(RandomListNode head) {
RandomListNode newHead = head.next;
while (head != null) {
RandomListNode temp = head.next;
head.next = temp.next;
head = head.next;
if (temp.next != null) {
temp.next = temp.next.next;
}
}
return newHead;
}
public RandomListNode copyRandomList(RandomListNode head) {
if (head == null) {
return null;
}
copyNext(head);
copyRandom(head);
return splitList(head);
}
}
164，interleaving---https://leetcode.com/problems/interleaving-string/---NOT BUG FREE
Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. For example, Given: s1 = "aabcc", s2 = "dbbca", When s3 = "aadbbcbcac", return true. When s3 = "aadbbbaccc", return false.
答案和思路：
public class Solution {
public boolean isInterleave(String s1, String s2, String s3) {
if (s1 == null || s2 == null || s3 == null) {
return false;
}
int len1 = s1.length();
int len2 = s2.length();
int len3 = s3.length();
if (len1 + len2 != len3) {
return false;
}
boolean[][] dp = new boolean[len1 + 1][len2 + 1];
dp[0][0] = true;
for (int i = 1; i < len1 + 1; i++) {
if (s1.charAt(i - 1) == s3.charAt(i - 1)) {
dp[i][0] = true;
} else {
break;
}
}
for (int i = 1; i < len2 + 1; i++) {
if (s2.charAt(i - 1) == s3.charAt(i - 1)) {
dp[0][i] = true;
} else {
break;
}
}
for (int i = 1; i < len1 + 1; i++) {
for (int j = 1; j < len2 + 1; j++) {
dp[i][j] = dp[i - 1][j] && s1.charAt(i - 1) == s3.charAt(i + j - 1) || dp[i][j - 1] && s2.charAt(j - 1) == s3.charAt(i + j - 1);
}
}
return dp[len1][len2];
}
}
163，word break---https://leetcode.com/problems/word-break/---NOT BUG FREE
Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words. For example, given s = "leetcode", dict = ["leet", "code"]. Return true because "leetcode" can be segmented as "leet code".
答案和思路：也就是这个能够划分的话取决于他前面的某一个能够划分并且dict包含了那个位置到他这里构成的字符串，为了优化，注意word的长度。
public class Solution {
public boolean wordBreak(String s, Set<String> wordDict) {
if (s == null || s.length() == 0) {
return true;
}
int len = s.length();
boolean[] dp = new boolean[len + 1];
dp[0] = true;
int wordMaxLen = 0;
for (String str : wordDict) {
wordMaxLen = Math.max(wordMaxLen, str.length());
}
for (int i = 1; i < len + 1; i++) {
for (int j = i - 1; j >= 0 && i - j <= wordMaxLen; j--) {
if (dp[j] && wordDict.contains(s.substring(j, i))) {
dp[i] = true;
break;
}
}
}
return dp[len];
}
}
162，jump gameII---https://leetcode.com/problems/jump-game-ii/---NOT BUG FREE
Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Your goal is to reach the last index in the minimum number of jumps. For example: Given array A = [2,3,1,1,4] The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.)
答案和思路：利用gready来做。
public class Solution {
public int jump(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int start = 0;
int end = 0;
int jumps = 0;
while (end < nums.length - 1) {
jumps++;
int farthest = 0;
for (int i = start; i <= end; i++) {
farthest = Math.max(farthest, nums[i] + i);
}
start = end;
end = farthest;
}
return jumps;
}
}
dp超时。
public int jump(int[] A) {
// write your code here
int[] dp = new int[A.length];
dp[0] = 0;
for (int i = 1; i < A.length; i++) {
int min = Integer.MAX_VALUE;
for (int j = 0; j < i; j++) {
if (A[j] >= i - j) {
min = Math.min(min, dp[j]);
}
}
dp[i] = min + 1;
}
return dp[A.length - 1];
}
161，jump game---https://leetcode.com/problems/jump-game/---NOT BUG FREE
Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Determine if you are able to reach the last index. For example: A = [2,3,1,1,4], return true. A = [3,2,1,0,4], return false.
答案和思路：greedy才能过。
public class Solution {
public boolean canJump(int[] nums) {
if (nums == null || nums.length == 0) {
return false;
}
int farthest = nums[0];
for (int i = 1; i < nums.length; i++) {
if (i <= farthest && nums[i] + i >= farthest) {
farthest = nums[i] + i;
if (farthest >= nums.length - 1) {
return true;
}
}
}
return farthest >= nums.length - 1;
}
}
用动态规划是过不了：
public class Solution {
public boolean canJump(int[] nums) {
// if (nums == null || nums.length == 0) {
// return true;
// }
boolean[] canJump = new boolean[nums.length];
canJump[0] = true;
for (int i = 0; i < nums.length; i++) {
if (!canJump[i]) {
return false;
}
for (int j = nums[i]; j >= 1; j--) {
if (i + j >= nums.length - 1) {
return true;
}
canJump[i + j] = true;
}
}
return canJump[nums.length - 1];
}
}
160，longest increasing subsequence---https://leetcode.com/problems/longest-increasing-subsequence/---BUG FREE
Given an unsorted array of integers, find the length of longest increasing subsequence. For example, Given [10, 9, 2, 5, 3, 7, 101, 18], The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length. Your algorithm should run in O(n2) complexity.
答案和思路：
public class Solution {
public int lengthOfLIS(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int result = 0;
int[] dp = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
dp[i] = 1;
for (int j = 0; j < i; j++) {
if (nums[j] < nums[i]) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
result = Math.max(result, dp[i]);
}
return result;
}
}
还有nlogn的解法。下一次来优化。
159,longest consecutive subsequence---https://leetcode.com/problems/longest-consecutive-sequence/---NOT BUG FREE
Given an unsorted array of integers, find the length of the longest consecutive elements sequence. For example, Given [100, 4, 200, 1, 3, 2], The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4. Your algorithm should run in O(n) complexity.
答案和思路：
public class Solution {
public int longestConsecutive(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int result = 0;
Set<Integer> set = new HashSet();
for (int i = 0; i < nums.length; i++) {
set.add(nums[i]);
}
for (int i = 0; i < nums.length; i++) {
int left = nums[i] - 1;
while (set.contains(left)) {
set.remove(left);
left--;
}
int right = nums[i] + 1;
while (set.contains(right)) {
set.remove(right);
right++;
}
result = Math.max(result, right - left - 1);
}
return result;
}
}
158，triangle---https://leetcode.com/problems/triangle/---NOT BUG FREE
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
if (null == triangle || triangle.size() == 0) {
return 0;
}
int min = Integer.MAX_VALUE;
int height = triangle.size();
for (int i = height - 2; i >= 0; i--) {
int size = triangle.get(i).size();
for (int j = 0; j < size; j++) {
triangle.get(i).set(j, triangle.get(i).get(j) + Math.min(triangle.get(i + 1).get(j), triangle.get(i + 1).get(j + 1)));
}
}
return triangle.get(0).get(0);
}
}
答案和思路：
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
if (null == triangle || triangle.size() == 0) {
return 0;
}
int min = Integer.MAX_VALUE;
int height = triangle.size();
for (int i = height - 2; i >= 0; i--) {
int size = triangle.get(i).size();
for (int j = 0; j < size; j++) {
triangle.get(i).set(j, triangle.get(i).get(j) + Math.min(triangle.get(i + 1).get(j), triangle.get(i + 1).get(j + 1)));
}
}
return triangle.get(0).get(0);
}
}
157,binary search tree iterator---https://leetcode.com/problems/binary-search-tree-iterator/---NOT BUG FREE
Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST. Calling next() will return the next smallest number in the BST. Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
答案和思路：
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BSTIterator {
Stack<TreeNode> stack = new Stack();
TreeNode cur;
public BSTIterator(TreeNode root) {
cur = root;
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return cur != null || !stack.isEmpty();
}
/** @return the next smallest number */
public int next() {
while (cur != null) {
stack.push(cur);
cur = cur.left;
}
cur = stack.pop();
TreeNode node = cur;
cur = cur.right;
return node.val;
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
156,inorder successor in binary search tree---http://www.lintcode.com/en/problem/inorder-successor-in-binary-search-tree/#---NOT BUG FREE
Given a binary search tree (See Definition) and a node in it, find the in-order successor of that node in the BST. If the given node has no in-order successor in the tree, return null. Notice It's guaranteed p is one node in the given tree. (You can directly compare the memory address to find p) Have you met this question in a real interview? Yes Example Given tree = [2,1] and node = 1: 2 / 1 return node 2. Given tree = [2,1,3] and node = 2: 2 / \ 1 3 return node 3.
答案和思路：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode inorderSuccessor(TreeNode root, TreeNode p) {
// write your code here
if (root == null) {
return null;
}
TreeNode successor = null;
while (root != null) {
if (root.val <= p.val) {
root = root.right;
} else {
successor = root;
root = root.left;
}
}
return successor;
}
}
155， binary tree maximum path sum---http://www.lintcode.com/en/problem/binary-tree-maximum-path-sum/---NOT BUG FREE
Given a binary tree, find the maximum path sum. The path may start and end at any node in the tree.
答案和思路：辅助一个函数是root走一边的最大path。那么对于所求，需要在考虑一下root.val + left + right这种情况。但是对于辅助函数，他只能返回走一边的max。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int maxPathSum(TreeNode root) {
if (root == null) {
return 0;
}
int[] max = {Integer.MIN_VALUE};
maxPathSum(root, max);
return max[0];
}
// 注意这里的辅助函数意思一定是root往一边走的最大路径。
private static int maxPathSum(TreeNode root, int[] max) {
if (root == null) {
return 0;
}
int leftSum = maxPathSum(root.left, max);
int rightSum = maxPathSum(root.right, max);
int curMax = root.val + Math.max(0, Math.max(leftSum, rightSum));
max[0] = Math.max(max[0], Math.max(curMax, root.val + leftSum + rightSum));
return curMax;
}
}
154,binary tree maximum path sum II(必须过根)---http://www.lintcode.com/en/problem/binary-tree-maximum-path-sum-ii/#---BUG FREE
Given a binary tree, find the maximum path sum from root. The path may end at any node in the tree and contain at least one node in it.
答案和思路：
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root the root of binary tree.
* @return an integer
*/
public int maxPathSum2(TreeNode root) {
// Write your code here
if (root == null) {
return 0;
}
return Math.max(root.val, root.val + Math.max(maxPathSum2(root.left), maxPathSum2(root.right)));
}
}
153，lowest common ancestor of a binary search tree---https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-search-tree/---NOT BUG FREE
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______6______
/ \
___2__ ___8__
/ \ / \
0 _4 7 9
/ \
3 5
For example, the lowest common ancestor (LCA) of nodes 2 and 8 is 6. Another example is LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.
答案和思路：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root.val < p.val && root.val < q.val) {
return lowestCommonAncestor(root.right, p, q);
} else if (root.val > p.val && root.val > q.val) {
return lowestCommonAncestor(root.left, p, q);
} else {
return root;
}
}
}
152,lowest common ancestor of a binary tree---https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/---BUG FREE
Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
7 4
For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
答案和思路：看先遇到谁。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null || p == root || q == root) {
return root;
}
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left != null && right != null) {
return root;
} else if (left != null) {
return left;
} else {
return right;
}
}
}
151,rotate string---http://www.lintcode.com/en/problem/rotate-string/---NOT BUG FREE
Given a string and an offset, rotate string by offset. (rotate from left to right)
答案和思路：三步旋转法。
public class Solution {
/**
* @param str: an array of char
* @param offset: an integer
* @return: nothing
*/
public void rotateString(char[] str, int offset) {
if (null == str || str.length == 0 || offset <= 0) {
return;
}
offset = offset % str.length;
reverseString(str, 0, str.length - 1 - offset);
reverseString(str, str.length - offset, str.length - 1);
reverseString(str, 0, str.length - 1);
}
public static void reverseString(char[] str, int start, int end) {
if (null == str || str.length == 0) {
return;
}
while (start < end) {
str[start] = (char) (str[start] ^ str[end]);
str[end] = (char) (str[start] ^ str[end]);
str[start] = (char) (str[start] ^ str[end]);
start++;
end--;
}
}
}
150, recover rotated sorted array---http://www.lintcode.com/en/problem/recover-rotated-sorted-array/---BUG FREE
Given a rotated sorted array, recover it to sorted array in-place.
答案和思路：三步恢复法。
public class Solution {
/**
* @param nums: The rotated sorted array
* @return: void
*/
public void recoverRotatedSortedArray(ArrayList<Integer> nums) {
// write your code
if (null == nums || nums.size() == 0) {
return;
}
for (int i = 0; i < nums.size() - 1; i++) {
if (nums.get(i) > nums.get(i + 1)) {
reverseArrayList(nums, 0, i);
reverseArrayList(nums, i + 1, nums.size() - 1);
reverseArrayList(nums, 0, nums.size() - 1);
}
}
}
// reverse the ArrayList
public static void reverseArrayList(ArrayList<Integer> a, int start, int end) {
// swap the a.get(start) and the a.get(end)
/** Error:
*while (start < end) {
* a.get(start) = a.get(start) ^ a.get(end);
* a.get(end) = a.get(start) ^ a.get(end);
* a.get(start) = a.get(start) ^ a.get(end);
* start++;
* end--;
* }
*/
while (start < end) {
a.set(start, a.get(start) ^ a.get(end));
a.set(end, a.get(start) ^ a.get(end));
a.set(start, a.get(start) ^ a.get(end));
start++;
end--;
}
}
}
149,在大数组中查找---http://www.lintcode.com/zh-cn/problem/search-in-a-big-sorted-array/---BUG FREE
给一个按照升序排序的正整数数组。这个数组很大以至于你只能通过固定的接口 ArrayReader.get(k) 来访问第k个数。(或者C++里是ArrayReader->get(k))，并且你也没有办法得知这个数组有多大。找到给出的整数target第一次出现的位置。你的算法需要在O(logk)的时间复杂度内完成，k为target第一次出现的位置的下标。 如果找不到target，返回-1。
答案和思路：
/**
* Definition of ArrayReader:
*
* class ArrayReader {
* // get the number at index, return -1 if not exists.
* public int get(int index);
* }
*/
public class Solution {
/**
* @param reader: An instance of ArrayReader.
* @param target: An integer
* @return : An integer which is the index of the target number
*/
public int searchBigSortedArray(ArrayReader reader, int target) {
// write your code here
if (reader == null) {
return -1;
}
int start = 0;
int end = Integer.MAX_VALUE;
int res = -1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (reader.get(mid) == -1) {
end = mid - 1;
continue;
}
if (reader.get(mid) == target) {
res = mid;
end = mid - 1;
} else if (reader.get(mid) < target) {
start = mid + 1;
} else {
end = mid - 1;
}
}
return res;
}
}
148, search for a range---https://leetcode.com/problems/search-for-a-range/---BUG FREE
Given a sorted array of integers, find the starting and ending position of a given target value. Your algorithm's runtime complexity must be in the order of O(log n). If the target is not found in the array, return [-1, -1]. For example, Given [5, 7, 7, 8, 8, 10] and target value 8, return [3, 4].
答案和思路：
public class Solution {
/**
*@param A : an integer sorted array
*@param target : an integer to be inserted
*return : a list of length 2, [index1, index2]
*/
public int[] searchRange(int[] a, int target) {
// write your code here
int[] res = new int[2];
res[0] = -1;
res[1] = -1;
if (a == null || a.length == 0) {
return res;
}
res[0] = findStartPosition(a, target);
res[1] = findEndPosition(a, target);
return res;
}
public static int findStartPosition(int[] a, int target) {
if (a == null || a.length == 0) {
return -1;
}
int start = 0;
int end = a.length - 1;
int res = -1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (a[mid] == target) {
res = mid;
end = mid - 1;
} else if (a[mid] < target) {
start = mid + 1;
} else {
end = mid - 1;
}
}
return res;
}
public static int findEndPosition(int[] a, int target) {
if (a == null || a.length == 0) {
return -1;
}
int start = 0;
int end = a.length - 1;
int res = -1;
while (start <= end) {
int mid = start + (end - start) / 2;
if (a[mid] == target) {
res = mid;
start = mid + 1;
} else if (a[mid] < target) {
start = mid + 1;
} else {
end = mid - 1;
}
}
return res;
}
}
147，find peak element---https://leetcode.com/problems/find-peak-element/---NOT BUG FREE
A peak element is an element that is greater than its neighbors. Given an input array where num[i] ≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is fine. You may imagine that num[-1] = num[n] = -∞. For example, in array [1, 2, 3, 1], 3 is a peak element and your function should return the index number 2.
答案和思路：注意，首先返回的是index，其次注意判断各种边界。只有一个元素，有两个元素的时候。
public class Solution {
public int findPeakElement(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
int left = 0;
int right = nums.length - 1;
while (left <= right) {
if (left == right) {
return left;
}
if (right - left == 1) {
return nums[right] > nums[left] ? right : left;
}
int mid = left + (right - left) / 2;
if (nums[mid - 1] < nums[mid] && nums[mid] > nums[mid + 1]) {
return mid;
} else if (nums[mid] < nums[mid + 1]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
}
146, first bad version---https://leetcode.com/problems/first-bad-version/---BUG FREE
You are a product manager and currently leading a team to develop a new product. Unfortunately, the latest version of your product fails the quality check. Since each version is developed based on the previous version, all the versions after a bad version are also bad. Suppose you have n versions [1, 2, ..., n] and you want to find out the first bad one, which causes all the following ones to be bad. You are given an API bool isBadVersion(version) which will return whether version is bad. Implement a function to find the first bad version. You should minimize the number of calls to the API.
答案和思路：
/* The isBadVersion API is defined in the parent class VersionControl.
boolean isBadVersion(int version); */
public class Solution extends VersionControl {
public int firstBadVersion(int n) {
if (n <= 0) {
return -1;
}
int left = 1;
int right = n;
int result = -1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (isBadVersion(mid)) {
result = mid;
right = mid - 1;
} else {
left = mid + 1;
}
}
return result;
}
}
145，search insert position---https://leetcode.com/problems/search-insert-position/---NOT BUG FREE
Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You may assume no duplicates in the array. Here are few examples. [1,3,5,6], 5 → 2 [1,3,5,6], 2 → 1 [1,3,5,6], 7 → 4 [1,3,5,6], 0 → 0
CUR:注意如果只有一个元素的时候该怎么考虑。
答案和思路：
public class Solution {
public int searchInsert(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int result = 0;
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] < target) {
result = mid + 1;
left = mid + 1;
} else {
right = mid -1;
}
}
return result;
}
}
144，last postition of target---http://www.lintcode.com/en/problem/last-position-of-target/#---BUG FREE
Find the last position of a target number in a sorted array. Return -1 if target does not exist.
答案和思路：
public class Solution {
/**
* @param nums: An integer array sorted in ascending order
* @param target: An integer
* @return an integer
*/
public int lastPosition(int[] nums, int target) {
// Write your code here
if (nums == null || nums.length == 0) {
return -1;
}
int left = 0;
int right = nums.length - 1;
int result = -1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
result = mid;
left = mid + 1;
} else if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return result;
}
}
143,first position of target---http://www.lintcode.com/en/problem/first-position-of-target/#---BUG FREE
For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity. If the target number does not exist in the array, return -1. Have you met this question in a real interview? Yes Example If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.
答案和思路：
class Solution {
/**
* @param nums: The integer array.
* @param target: Target to find.
* @return: The first position of target. Position starts from 0.
*/
public int binarySearch(int[] nums, int target) {
//write your code here
if (nums == null || nums.length == 0) {
return -1;
}
int result = -1;
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
result = mid;
right = mid - 1;
} else if (nums[mid] < target) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return result;
}
}
142,binary search.---BUG FREE
class Solution {
public static void main(String[] args) {
int[] nums = {1, 2, 5, 6};
System.out.println(binarySearch(nums, 2));
}
private static int binarySearch(int[] nums, int t) {
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == t) {
return mid;
} else if (nums[mid] > t) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return -1;
}
}
141，posts---
import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;
/**
* Created by Nano on 2016/9/29.
*/
public class Solution {
static String[] paginate(int num, String[] results) {
int length = results.length;
int count = length / num;
String[] output = new String[count + length];
boolean[] flag = new boolean[length];
;
; i <= count; i++) {
Set<String> keep = new HashSet();
; k < length && keep.size() < num; k++) {
int location = results[k].indexOf(",");
String ID = results[k].substring(, location);
if (!keep.contains(ID) && !flag[k]) {
keep.add(ID);
flag[k] = true;
output[start++] = results[k];
}
}
if (i != count && keep.size() == num) {
output[start++] = "";
} else {
int size = keep.size();
; size < num && k < length; k++) {
if (!flag[k]) {
flag[k] = true;
output[start++] = results[k];
size++;
}
}
if (i != count) {
output[start++] = "";
}
}
}
return output;
}
public static void main(String[] args) {
String[] results = {
"131,28,300.6,San Francisco",
"4,5,209.1,San Francisco",
"20,7,203.4,Oakland",
"6,8,202.9,San Francisco",
"6,10,199.8,San Francisco",
"1,16,190.5,San Francisco",
"6,29,185.3,San Francisco",
"7,20,180.0,Oakland",
"6,21,162.2,San Francisco",
"2,18,161.7,San Jose",
"2,30,149.8,San Jose",
"3,76,146.7,San Francisco",
"2,14,141.8,San Jose"
};
System., results)));
}
}
140, subsetsII ----https://leetcode.com/problems/subsets-ii/---BUG FREE
Given a collection of integers that might contain duplicates, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If nums = [1,2,2], a solution is: [ [2], [1], [1,2,2], [2,2], [1,2], [] ]
答案和思路:
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> results = new ArrayList();
List<Integer> cur = new ArrayList();
Arrays.sort(nums);
results.add(cur);
if (nums == null) {
return results;
}
for (int i = 0; i < nums.length; i++) {
List<List<Integer>> temp = new ArrayList(results);
for (List<Integer> list : results) {
List<Integer> listTemp = new ArrayList(list);
listTemp.add(nums[i]);
if (!temp.contains(listTemp)) {
temp.add(listTemp);
}
}
results = new ArrayList(temp);
}
return results;
}
}
递归
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> results = new ArrayList();
if (nums == null) {
return results;
}
Arrays.sort(nums);
List<Integer> cur = new ArrayList();
subsets(nums, 0, results, cur);
return results;
}
private static void subsets(int[] nums, int index, List<List<Integer>> results, List<Integer> cur) {
results.add(new ArrayList(cur));
for (int i = index; i < nums.length; i++) {
if (i != index && nums[i] == nums[i - 1]) {
continue;
}
cur.add(nums[i]);
subsets(nums, i + 1, results, cur);
cur.remove(cur.size() - 1);
}
}
}
139,subsets---https://leetcode.com/problems/subsets/--- BUG FREE
Given a set of distinct integers, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If nums = [1,2,3], a solution is: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
答案和思路：
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> results = new ArrayList();
List<Integer> cur = new ArrayList();
results.add(cur);
if (nums == null) {
return results;
}
for (int i = 0; i < nums.length; i++) {
List<List<Integer>> temp = new ArrayList(results);
for (List<Integer> list : results) {
List<Integer> listTemp = new ArrayList(list);
listTemp.add(nums[i]);
temp.add(listTemp);
}
results = new ArrayList(temp);
}
return results;
}
}
递归：
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> results = new ArrayList();
if (nums == null) {
return results;
}
List<Integer> cur = new ArrayList();
subsets(nums, 0, results, cur);
return results;
}
private static void subsets(int[] nums, int index, List<List<Integer>> results, List<Integer> cur) {
results.add(new ArrayList(cur));
for (int i = index; i < nums.length; i++) {
cur.add(nums[i]);
subsets(nums, i + 1, results, cur);
cur.remove(cur.size() - 1);
}
}
}
138,strStr(找子串的起点)---https://leetcode.com/problems/implement-strstr/---BUG FREE
Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
答案和思路：只需要暴力扫一遍就行。注意的是边界问题，对于大的string他的开始只能够从他的长度减去小string的长度。
public class Solution {
public int strStr(String haystack, String needle) {
if (haystack == null || needle == null) {
return -1;
}
for (int i = 0; i < haystack.length() - needle.length() + 1; i++) {
int j = 0;
for (; j < needle.length(); j++) {
if (haystack.charAt(i + j) != needle.charAt(j)) {
break;
}
}
if (j == needle.length()) {
return i;
}
}
return -1;
}
}
137,isomorphic strings---https://leetcode.com/problems/isomorphic-strings/---NOT BUG FREE
Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. All occurrences of a character must be replaced with another character while preserving the order of characters. No two characters may map to the same character but a character may map to itself. For example, Given "egg", "add", return true. Given "foo", "bar", return false. Given "paper", "title", return true.
答案和思路：利用map的put返回值，但是注意需要双map。
public class Solution {
public boolean isIsomorphic(String s, String t) {
if (s == null || t == null) {
return true;
} else if (s == null) {
return false;
} else if (t == null || s.length() != t.length()) {
return false;
}
Map map1 = new HashMap();
Map map2 = new HashMap();
for (int i = 0; i < s.length(); i++) {
if (!Objects.equals(map1.put(s.charAt(i), i), map2.put(t.charAt(i), i))) {
return false;
}
}
return true;
}
}
136,clone graph---https://leetcode.com/problems/clone-graph/---NOT BUG FREE
Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors.
OJ's undirected graph serialization:
Nodes are labeled uniquely.
We use # as a separator for each node, and , as a separator for node label and each neighbor of the node.
As an example, consider the serialized graph {0,1,2#1,2#2,2}.
The graph has a total of three nodes, and therefore contains three parts as separated by #.
First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
Second node is labeled as 1. Connect node 1 to node 2.
Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle.
Visually, the graph looks like the following:
1
/ \
/ \
0 --- 2
/ \
\_/
答案和思路：利用map来对应新旧表。用queue来记录node。
/**
* Definition for undirected graph.
* class UndirectedGraphNode {
* int label;
* List<UndirectedGraphNode> neighbors;
* UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
* };
*/
public class Solution {
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) {
return null;
}
UndirectedGraphNode head = new UndirectedGraphNode(node.label);
Queue<UndirectedGraphNode> queue = new LinkedList();
queue.add(node);
Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap();
map.put(node, head);
while (!queue.isEmpty()) {
UndirectedGraphNode curNode = queue.poll();
for (UndirectedGraphNode now : curNode.neighbors) {
if (!map.containsKey(now)) {
map.put(now, new UndirectedGraphNode(now.label));
queue.add(now);
}
map.get(curNode).neighbors.add(map.get(now));
}
}
return head;
}
}
135,Group Anagrams---https://leetcode.com/problems/anagrams/---NOT BUG FREE
Given an array of strings, group anagrams together. For example, given: ["eat", "tea", "tan", "ate", "nat", "bat"], Return: [ ["ate", "eat","tea"], ["nat","tan"], ["bat"] ]
答案和思路：注意，map的key一定是可以比较的。一开始写char[]就是错的。
public class Solution {
public static List<List<String>> groupAnagrams(String[] strs) {
List<List<String>> result = new ArrayList();
if (strs == null || strs.length == 0) {
return result;
}
Map<String, List<String>> map = new HashMap();
for (int i = 0; i < strs.length; i++) {
char[] now = strs[i].toCharArray();
Arrays.sort(now);
String str = new String(now);
if (map.containsKey(str)) {
List<String> cur = map.get(str);
cur.add(strs[i]);
map.put(str, cur);
} else {
List<String> cur = new ArrayList();
cur.add(strs[i]);
map.put(str, cur);
}
}
for (Map.Entry<String, List<String>> entry : map.entrySet()) {
result.add(entry.getValue());
}
return result;
}
}
134,word pattern---https://leetcode.com/problems/word-pattern/---NOT BUG FREE
Given a pattern and a string str, find if str follows the same pattern. Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str. Examples: pattern = "abba", str = "dog cat cat dog" should return true. pattern = "abba", str = "dog cat cat fish" should return false. pattern = "aaaa", str = "dog cat cat dog" should return false. pattern = "abba", str = "dog dog dog dog" should return false. Notes: You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.
答案和思路：注意这个匹配是一一对应的。所以不仅要对的过去，还要对得回来。所以思路利用：一直往里加，如果一个已经重复了，但是另一个还没有重复。那么说明一一对应不上了。
上面这句话有错误。之所以用set不行，而用map行，是因为map.put的返回值是他被覆盖掉的那个值（没有的时候就是null）。
public class Solution {
public boolean wordPattern(String pattern, String str) {
//注意，map.put（）返回的是他代替掉的值，如果还没有存入，那么就是null。所以他可以用来判断是否一致。
if (pattern == null && str == null) {
return true;
} else if (pattern == null || str == null) {
return false;
}
String[] words = str.split(" ");
if (pattern.length() != words.length) {
return false;
}
Map map = new HashMap();
for (int i = 0; i < pattern.length(); i++) {
if (!Objects.equals(map.put(words[i], i), map.put(pattern.charAt(i), i))) {
return false;
}
}
return true;
}
}
133,two sum---https://leetcode.com/problems/two-sum/---BUG FREE
Given an array of integers, return indices of the two numbers such that they add up to a specific target. You may assume that each input would have exactly one solution. Example: Given nums = [2, 7, 11, 15], target = 9, Because nums[0] + nums[1] = 2 + 7 = 9, return [0, 1].
答案和思路：
public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] results = new int[2];
if (nums == null || nums.length == 0) {
return results;
}
Map<Integer, Integer> map = new HashMap();
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(nums[i])) {
results[0] = map.get(nums[i]);
results[1] = i;
return results;
} else {
map.put(target - nums[i], i);
}
}
return results;
}
}
132,map，set的基础知识
map:
HashMap:无序;TreeMap:有序。LinkedHashMap:插入顺序。
set : ArrayList : 有重复的。 HashSet：无序。TreeSet：有序。LinkedHashSet：插入顺序。
add, remove, get, containsKey都是O（1）。containsValue是O（n）；
map的遍历：
public static void main(String[] args) {
Map<Integer, Integer> map = new HashMap();
map.put(1, 2);
for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
System.out.println(entry.getKey());
}
}
131-快排---NOT BUG FREE
import java.util.Arrays;
import java.util.Collections;
import java.util.PriorityQueue;
public class Kolmostar {
public static void main(String[] args) {
int[] a = {1, 7, 16, 7, 9};
quickSort(a, 0, 4);
System.out.println(Arrays.toString(a));
}
private static void quickSort(int[] a, int left, int right) {
if (left >= right) {
return;
}
int first = left;
int last = right;
int key = a[first];
while (first < last) {
while (first < last && a[last] >= key) {
last--;
}
a[first] = a[last];
while (first < last && a[first] <= key) {
first++;
}
a[last] = a[first];
}
a[first] = key;
quickSort(a, left, first - 1);
quickSort(a, first + 1, right);
}
}
130，kth largest element in array---https://leetcode.com/problems/kth-largest-element-in-an-array/---NOT BUG FREE
Find the kth largest element in an unsorted array. Note that it is the kth largest element in the sorted order, not the kth distinct element. For example, Given [3,2,1,5,6,4] and k = 2, return 5.
答案和思路：用快排之所以复杂度是O（n）是因为，第一个partition是O(N),第二次是O(N/2).n + n/2 + n/4 == 2n
public class Solution {
public int findKthLargest(int[] nums, int k) {
if (nums == null || nums.length == 0 || k <= 0) {
return 0;
}
return helper(nums, 0, nums.length - 1, nums.length - k + 1);
}
private static int helper(int[] nums, int left, int right, int k) {
if (left == right) {
return nums[left];
}
int position = partition(nums, left, right);
if (position + 1 == k) {
return nums[position];
} else if (position + 1 < k) {
return helper(nums, position + 1, right, k);
} else {
return helper(nums, left, position - 1, k);
}
}
private static int partition(int[] nums, int left, int right) {
int first = left;
int last = right;
int key = nums[first];
while (first < last) {
while (first < last && nums[last] >= key) {
last--;
}
nums[first] = nums[last];
while (first < last && nums[first] <= key) {
first++;
}
nums[last] = nums[first];
}
nums[first] = key;
return first;
}
}
129,merge k sorted list---https://leetcode.com/problems/merge-k-sorted-lists/---NOT BUG FREE
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
CUR:思路不清晰
答案和思路：注意要利用好每个list是有序的。minHeap里面存的是node。
public class Solution {
public ListNode mergeKLists(ListNode[] lists) {
ListNode newHead = new ListNode(-1);
// PriorityQueue<Integer> minHeap = new PriorityQueue();
PriorityQueue<ListNode> minHeap = new PriorityQueue(new Comparator<ListNode>() {
public int compare(ListNode a, ListNode b) {
return a.val - b.val;
}
});
for (int i = 0; i < lists.length; i++) {
if (lists[i] != null) {
minHeap.add(lists[i]);
}
}
ListNode cur = newHead;
while (!minHeap.isEmpty()) {
cur.next = minHeap.poll();
cur = cur.next;
if (cur.next != null) {
minHeap.add(cur.next);
}
}
return newHead.next;
}
}
存val。
public class Solution {
public ListNode mergeKLists(ListNode[] lists) {
ListNode newHead = new ListNode(-1);
PriorityQueue<Integer> minHeap = new PriorityQueue();
for (int i = 0; i < lists.length; i++) {
ListNode node = lists[i];
while (node != null) {
minHeap.add(node.val);
node = node.next;
}
}
ListNode cur = newHead;
while (!minHeap.isEmpty()) {
cur.next = new ListNode(minHeap.poll());
cur = cur.next;
}
return newHead.next;
}
}
128，merge sorted list---https://leetcode.com/problems/merge-two-sorted-lists/---BUG FREE
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
答案和思路：注意返回的是newHead.next;
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
ListNode newHead = new ListNode(-1);
ListNode node1 = l1;
ListNode node2 = l2;
ListNode node = newHead;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
node.next = new ListNode(l1.val);
l1 = l1.next;
} else {
node.next = new ListNode(l2.val);
l2 = l2.next;
}
node = node.next;
}
while (l1 != null) {
node.next = new ListNode(l1.val);
l1 = l1.next;
node = node.next;
}
while (l2 != null) {
node.next = new ListNode(l2.val);
l2 = l2.next;
node = node.next;
}
return newHead.next;
}
}
127,merge sorted array---https://leetcode.com/problems/merge-sorted-array/---BUG FREE
Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array. Note: You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2. The number of elements initialized in nums1 and nums2 are m and n respectively.
答案和思路：
public class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int index = m + n - 1;
while (m > 0 && n > 0) {
if (nums1[m - 1] > nums2[n - 1]) {
nums1[index] = nums1[m - 1];
m--;
} else {
nums1[index] = nums2[n - 1];
n--;
}
index--;
}
while (m > 0) {
nums1[index] = nums1[m - 1];
index--;
m--;
}
while (n > 0) {
nums1[index] = nums2[n - 1];
index--;
n--;
}
}
}
126，heap的知识。--复杂度分析。
add/insert:O(n)。把数字插入到最后一个位置，然后不断地往上走。 delete.删除。把最后一个和顶元素交换。然后把他不断地往下走。 建堆：1，如果是一个接一个的加入。复杂度是：nlogn. 2，但是通常的建堆是把堆建好。从倒数第二层开始一层一层的往下走。那么递推公式就是： 看似建堆(make_heap)的复杂度为O(nlogn)，实则为O(n)，reheapify_down不会touch到每个节点，所以不算简单的递归，只能算叠加。证明如下： 因为堆构建的是一颗平衡二叉树。于是有： 1. 一棵拥有n个节点的平衡二叉树，树高 h 约为 lg n 。 2. 树的第 i 层 最多有节点 2^i 个。注意，第 i 层的高度为 i + 1。如第0层的高度为1，拥有1个根节点。 那么做reheapify_down时，最底层(h-1)的节点不会动，倒数第二层（h-2）的节点最多往下移动一次，倒数第三层（h-3）的节点最多往下移动2次....第0层的节点往下最多移动（h-1）次。 所以，最坏情况下，每层所有节点都会往下移到最底层。 则，所有操作总和为 S = 2^(h-1)*0 + 2^(h-2)*1 + 2^(h-3) * 2 + ... + 2^1*(h-2) + 2^0*(h-1) ----- （1） 把（1）式乘以2，再减去（1）式， 可得 S = 2^(h-1) + 2^(h-2) + ... + 2^1 - 2^0*(h-1) = 2(1-2^(h-1))/(1-2) - (h-1) = 2^h - h- 1 ---- (2) 把h = lg n 代入 （2）式， 得 S = n - lgn - 1 <= n (n >=1) 故而， 建堆复杂度为O(n) 。
125,course schedule---https://leetcode.com/problems/course-schedule/---NOT BUG FREE
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1] Given the total number of courses and a list of prerequisite pairs, is it possible for you to finish all courses? For example: 2, [[1,0]] There are a total of 2 courses to take. To take course 1 you should have finished course 0. So it is possible. 2, [[1,0],[0,1]] There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
CUR:思路不清晰
答案和思路：
public class Solution {
public boolean canFinish(int numCourses, int[][] prerequisites) {
List<List<Integer>> graph = new ArrayList();
for (int i = 0; i < numCourses; i++) {
graph.add(new ArrayList<Integer>());
}
for (int i = 0; i < prerequisites.length; i++) {
graph.get(prerequisites[i][1]).add(prerequisites[i][0]);
}
int[] status = new int[numCourses];
for (int i = 0; i < numCourses; i++) {
if (status[i] == 0 && !isDAG(graph, i, status)) {
1 return false;
}
}
return true;
}
private static boolean isDAG(List<List<Integer>> graph, int root, int[] status) {
if (status[root] == 1) {
return false;
}
status[root] = 1;
for (int i : graph.get(root)) {
if (!isDAG(graph, i, status)) {
return false;
}
}
status[root] = 2;
return true;
}
}
124, binary tree zigzag level order traversal---https://leetcode.com/problems/binary-tree-zigzag-level-order-traversal/---BUG FREE
Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
答案和思路：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList();
if (root == null) {
return result;
}
Queue<TreeNode> queue = new LinkedList();
queue.add(root);
boolean flag = true;
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> cur = new ArrayList();
for (int i = 0; i < size; i++) {
TreeNode now = queue.poll();
if (flag) {
cur.add(now.val);
} else {
cur.add(0, now.val);
}
if (now.left != null) {
queue.add(now.left);
}
if (now.right != null) {
queue.add(now.right);
}
}
result.add(new ArrayList(cur));
flag = !flag;
}
return result;
}
}
123，binary-tree-level-order-traversal---https://leetcode.com/problems/binary-tree-level-order-traversal/---BUG FREE
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
答案和思路：利用queue，但是size要单独存下来。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList();
if (root == null) {
return result;
}
Queue<TreeNode> queue = new LinkedList();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> cur = new ArrayList();
for (int i = 0; i < size; i++) {
TreeNode now = queue.poll();
cur.add(now.val);
if (now.left != null) {
queue.offer(now.left);
}
if (now.right != null) {
queue.offer(now.right);
}
}
result.add(new ArrayList(cur));
}
return result;
}
}
122，flatten binary tree to lined list---https://leetcode.com/problems/flatten-binary-tree-to-linked-list/---NOT BUG FREE
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1
/ \
2 5
/ \ \
3 4 6
The flattened tree should look like:
1
\
2
\
3
\
4
\
5
\
6
CUR：思路不清晰。
答案和思路：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
TreeNode lastNode = null;
public void flatten(TreeNode root) {
if (root == null) {
return;
}
if (lastNode != null) {
lastNode.left = null;
lastNode.right = root;
}
lastNode = root;
TreeNode right = root.right;
flatten(root.left);
flatten(right);
}
}
121,validate binary search tree---https://leetcode.com/problems/validate-binary-search-tree/--BUG FREE
Given a binary tree, determine if it is a valid binary search tree (BST).
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node's key.
The right subtree of a node contains only nodes with keys greater than the node's key.
Both the left and right subtrees must also be binary search trees.
Example 1:
2
/ \
1 3
Binary tree [2,1,3], return true.
Example 2:
1
/ \
2 3
Binary tree [1,2,3], return false.
CUR:可能会给整数。所以需要用Long.MAX_VALUE来做。
答案和思路：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isValidBST(TreeNode root) {
if (root == null) {
return true;
}
return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
}
private static boolean isValidBST(TreeNode root, long left, long right) {
if (root == null) {
return true;
}
if (root.val <= left || root.val >= right) {
return false;
} else {
return isValidBST(root.left, left, root.val) && isValidBST(root.right, root.val, right);
}
}
}
120,Same Tree---https://leetcode.com/problems/same-tree/---BUG FREE
Given two binary trees, write a function to check if they are equal or not. Two binary trees are considered equal if they are structurally identical and the nodes have the same value.
答案和思路：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null && q == null) {
return true;
} else if (p == null || q == null || p.val != q.val) {
return false;
}
return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
}
}
119,palindrome partition---https://leetcode.com/problems/palindrome-partitioning/---NOT BUG FREE
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab", Return [ ["aa","b"], ["a","a","b"] ]
CUR:思路不清晰
答案和思路：记住走到了哪一个index。
public class Solution {
public List<List<String>> partition(String s) {
List<List<String>> results = new ArrayList();
if (s == null || s.length() == 0) {
return results;
}
ArrayList cur = new ArrayList();
partition(results, s, 0, cur);
return results;
}
private static void partition(List<List<String>> results, String s, int start, List<String> path) {
if (start == s.length()) {
results.add(new ArrayList(path));
return;
}
for (int i = start + 1; i <= s.length(); i++) {
String sub = s.substring(start, i);
if (isPalindrome(sub)) {
path.add(sub);
partition(results, s, i, path);
path.remove(path.size() - 1);
}
}
}
private static boolean isPalindrome(String s) {
for (int i = 0, j = s.length() - 1; i < j; i++, j--) {
if (s.charAt(i) != s.charAt(j)) {
return false;
}
}
return true;
}
}
118,----https://leetcode.com/problems/word-search/---NOT BUG FREE
Given a 2D board and a word, find if the word exists in the grid. The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once. For example, Given board = [ ['A','B','C','E'], ['S','F','C','S'], ['A','D','E','E'] ] word = "ABCCED", -> returns true, word = "SEE", -> returns true, word = "ABCB", -> returns false.
CUR：思路不清晰
答案和思路：就是看从这一个点开始走，是否能够找到。
public class Solution {
public boolean exist(char[][] board, String word) {
if (board.length == 0 || board[0].length == 0) {
return false;
}
boolean[][] flag = new boolean[board.length][board[0].length];
for (int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
if (exist(board, i, j, word, 0, flag)) {
return true;
}
}
}
return false;
}
private static boolean exist(char[][] board, int x, int y, String word, int index, boolean[][] flag) {
if (index == word.length()) {
return true;
}
if (x < 0 || x >= board.length || y < 0 || y >= board[0].length || flag[x][y] || board[x][y] != word.charAt(index)) {
return false;
}
int[] dx = {1, 0, -1, 0};
int[] dy = {0, 1, 0, -1};
for (int i = 0; i < 4; i++) {
flag[x][y] = true;
if (exist(board, x + dx[i], y + dy[i], word, index + 1, flag)) {
return true;
}
flag[x][y] = false;
}
return false;
}
}
117，---https://leetcode.com/problems/find-median-from-data-stream/---NOT BUG FREE
Median is the middle value in an ordered integer list. If the size of the list is even, there is no middle value. So the median is the mean of the two middle value. Examples: [2,3,4] , the median is 3 [2,3], the median is (2 + 3) / 2 = 2.5 Design a data structure that supports the following two operations: void addNum(int num) - Add a integer number from the data stream to the data structure. double findMedian() - Return the median of all elements so far. For example: add(1) add(2) findMedian() -> 1.5 add(3) findMedian() -> 2
答案和思路：中位数位于的位置就是大于前半部分，小于后半部分。也就是小的里面的最大（小的放大顶堆，顶就是最大的），大的里面的最小（大的放小顶堆，堆顶就是最小）。
public class MedianFinder {
PriorityQueue<Integer> minHeap = new PriorityQueue();
PriorityQueue<Integer> maxHeap = new PriorityQueue(Collections.reverseOrder());
// Adds a number into the data structure.
public void addNum(int num) {
maxHeap.add(num);
minHeap.add(maxHeap.poll());
if (maxHeap.size() < minHeap.size()) {
maxHeap.add(minHeap.poll());
}
}
// Returns the median of current data stream
public double findMedian() {
if (maxHeap.size() == minHeap.size()) {
return (maxHeap.peek() + minHeap.peek()) / 2.0;
} else {
return maxHeap.peek();
}
}
};
// Your MedianFinder object will be instantiated and called as such:
// MedianFinder mf = new MedianFinder();
// mf.addNum(1);
// mf.findMedian();
116,---median of two sorted arrays---https://leetcode.com/problems/median-of-two-sorted-arrays/---NOT BUG FREE
There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). Example 1: nums1 = [1, 3] nums2 = [2] The median is 2.0 Example 2: nums1 = [1, 2] nums2 = [3, 4] The median is (2 + 3)/2 = 2.5
CUR：思路不清晰。k == 1是关键的返回。
答案和思路：利用查找第k大的数字。相对较小的那个中点的前半部分是一定可以排除的。k指的是第k大，而不是坐标。
public class Solution {
public double findMedianSortedArrays(int[] nums1, int[] nums2) {
if (nums1 == null || nums2 == null) {
return -1;
}
int len1 = nums1.length;
int len2 = nums2.length;
int len = len1 + len2;
if (len % 2 == 1) {
return findMedian(nums1, 0, nums2, 0, len / 2 + 1);
} else {
return (findMedian(nums1, 0, nums2, 0, len / 2) + findMedian(nums1, 0, nums2, 0, len / 2 + 1)) / 2.0;
}
}
private static double findMedian(int[] a, int aIndex, int[] b, int bIndex, int k) {
if (aIndex >= a.length) {
return b[bIndex + k - 1];
}
if (bIndex >= b.length) {
return a[aIndex + k - 1];
}
if (k == 1) {
return Math.min(a[aIndex], b[bIndex]);
}
int aMid = aIndex + k / 2 - 1 < a.length ? a[aIndex + k / 2 - 1] : Integer.MAX_VALUE;
int bMid = bIndex + k / 2 - 1 < b.length ? b[bIndex + k / 2 - 1] : Integer.MAX_VALUE;
if (aMid < bMid) {
return findMedian(a, aIndex + k / 2, b, bIndex, k - k / 2);
} else {
return findMedian(a, aIndex, b, bIndex + k / 2, k - k / 2);
}
}
}
115,---median of two sorted array---找两个数组的中位数（有序无序）
因为给的条件都比较独特，为了弥补早上发挥的不足，所以，写了早上面试过程中讨论的三种情况的答案。
1，师兄给出的题目：给两个数组，都是长度为n。需要找的中位数是返回C(n)（C代表把两个数组合并在一起了）.所以，这里面说明了需要实际查找的是第N + 1位。
根据这个背景下，给出代码如下：
public class Kolmostar {
/**
* @author 李春跃-北京邮电大学研究生
* @email chunyuelibigdata@gmail.com
* @思路 利用二分来做。其实早上说的时候基本上思路已经正确了，就是比较中间的值，然后往哪边走。
* 当时的说法错误在于不应该两个都动，而是动其中较小的那个就行。排除掉那部分肯定不会含有值的地方。
* @分析 时间复杂度O(log(n)).空间复杂度O(1)
*/
public static void main(String[] args) {
int[] A = {1, 6, 7, 9};
int[] B = {0, 2, 4, 5};
System.out.println(findMedian(A, B));
}
private static int findMedian(int A[], int B[]) {
if (A == null || B == null) {
return -1; // 判断是否有异常，返回-1.
}
int n = A.length; //早上题目描述是给两个都是长度为n的数组。返回的是C(n)。那么也就是找第n+1位.
// int len = A.length + B.length;
// if (len % 2 == 1) {
// return findKth(A, 0, B, 0, len / 2 + 1);
// }
// return (
// findKth(A, 0, B, 0, len / 2) + findKth(A, 0, B, 0, len / 2 + 1)
// ) / 2.0;
return findKth(A, 0, B, 0, n + 1);
}
// find kth number of two sorted array
private static int findKth(int[] A, int A_index, int[] B, int B_index, int k){
if (A_index >= A.length) {
return B[B_index + k - 1];
}
if (B_index >= B.length) {
return A[A_index + k - 1];
}
if (k == 1) {
return Math.min(A[A_index], B[B_index]);
}
int A_key = A_index + k / 2 - 1 < A.length ? A[A_index + k / 2 - 1] : Integer.MAX_VALUE;
int B_key = B_index + k / 2 - 1 < B.length ? B[B_index + k / 2 - 1]: Integer.MAX_VALUE;
if (A_key < B_key) {
return findKth(A, A_index + k / 2, B, B_index, k - k / 2);
} else {
return findKth(A, A_index, B, B_index + k / 2, k - k / 2);
}
}
}
2，因为严格意义上这道题给出的两个数组不一定都是一样长为n，而且中位数的严格定义应该是总个数为奇数的时候是最中间那个，偶数个的时候是最中间的那两个的均值。所以，相对于上面的这里需要作出以下几个改变：
（1）返回值应该是double而不是整数。
（2）需要分情况讨论总个数是奇数还是偶数的时候。
（3）时间复杂度变成了O(log(Math. min(A.length, B.length)))。
在这个背景下，给出代码：
public class Kolmostar {
/**
* @author 李春跃-北京邮电大学研究生
* @email chunyuelibigdata@gmail.com
* @思路 利用能够找到第k大的函数来分个数为奇数偶数情况的时候求出中位数
* @时间复杂度O(log(Math. min(A.length, B.length)))
* @空间复杂度O(1)
*/
public static void main(String[] args) {
int[] A = {1, 6, 7, 9};
int[] B = {0, 2, 4, 5};
System.out.println(findMedian(A, B));
}
private static double findMedian(int A[], int B[]) {
if (A == null || B == null) {
return -1; // 判断是否有异常，返回-1.
}
int len = A.length + B.length;
if (len % 2 == 1) {
return findKth(A, 0, B, 0, len / 2 + 1);
}
return (findKth(A, 0, B, 0, len / 2) + findKth(A, 0, B, 0, len / 2 + 1)) / 2.0;
}
// find kth number of two sorted array
private static int findKth(int[] A, int A_index, int[] B, int B_index, int k){
if (A_index >= A.length) {
return B[B_index + k - 1];
}
if (B_index >= B.length) {
return A[A_index + k - 1];
}
if (k == 1) {
return Math.min(A[A_index], B[B_index]);
}
int A_key = A_index + k / 2 - 1 < A.length ? A[A_index + k / 2 - 1] : Integer.MAX_VALUE;
int B_key = B_index + k / 2 - 1 < B.length ? B[B_index + k / 2 - 1]: Integer.MAX_VALUE;
if (A_key < B_key) {
return findKth(A, A_index + k / 2, B, B_index, k - k / 2);
} else {
return findKth(A, A_index, B, B_index + k / 2, k - k / 2);
}
}
}
3，早上面试的时候，也讨论到了如果两个数组没有序的话该怎么做。回来学校也写了这种情况的代码。
import java.util.PriorityQueue;
public class Kolmostar {
/**
* @author 李春跃-北京邮电大学研究生
* @email chunyuelibigdata@gmail.com
* @思路 利用最大堆最小堆来辅助一起维持中位数的位置
* @时间复杂度：O(A.length + B.length)
* @空间复杂度：O(A.length + B.length)
*/
public static void main(String[] args) {
int[] A = {1, 6, 7, 9};
int[] B = {0, 2, 4, 5};
System.out.println(findMedian(A, B));
}
public static double findMedian(int[] A, int[] B) {
if (A == null || B == null) {
return -1;
}
PriorityQueue<Integer> minHeap = new PriorityQueue();
PriorityQueue<Integer> maxHeap = new PriorityQueue();
for (int i = 0; i < A.length; i++) {
minHeap.offer(A[i]);
maxHeap.offer(-minHeap.poll());
if (maxHeap.size() - minHeap.size() > 1) {
minHeap.offer(-maxHeap.poll());
}
}
for (int i = 0; i < B.length; i++) {
minHeap.offer(B[i]);
maxHeap.offer(-minHeap.poll());
if (maxHeap.size() - minHeap.size() > 1) {
minHeap.offer(-maxHeap.poll());
}
}
//返回的时候，讨论总个数是偶数还是奇数
if (minHeap.size() == maxHeap.size()) {
return (double)(minHeap.peek() - maxHeap.peek()) / 2.0;
}
return -maxHeap.peek();
}
}
----by 李春跃
114，---maximum path sum---https://leetcode.com/problems/binary-tree-maximum-path-sum/---NOT BUG FREE
Given a binary tree, find the maximum path sum.
For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root.
For example:
Given the below binary tree,
1
/ \
2 3
Return 6.
CUR:思路不清晰
答案和思路：单写一个函数，计算的是过这个点的max。那么过每一个点的max里面的最大就是所求。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int maxPathSum(TreeNode root) {
if (root == null) {
return 0;
}
int[] max = new int[1];
max[0] = Integer.MIN_VALUE;
maxPathSum(root, max);
return max[0];
}
private static int maxPathSum(TreeNode root, int[] max) {
if (root == null) {
return 0;
}
int leftSum = maxPathSum(root.left, max);
int rightSum = maxPathSum(root.right, max);
int curMax = root.val + Math.max(0, Math.max(leftSum, rightSum));
max[0] = Math.max(max[0], Math.max(curMax, leftSum + root.val + rightSum));
return curMax;
}
}
113, path-sum II---https://leetcode.com/problems/path-sum-ii/---NOT BUG FREE
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
CUR:思路不清晰。
答案和思路：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> results = new ArrayList();
if (root == null) {
return results;
}
List<Integer> cur = new ArrayList();
pathSum(root, sum, results, cur);
return results;
}
private static void pathSum(TreeNode root, int sum, List<List<Integer>> results, List<Integer> cur) {
if (root == null) {
return;
}
sum = sum - root.val;
if (root.left == null && root.right == null) {
if (sum == 0) {
cur.add(root.val);
results.add(new ArrayList<Integer>(cur));
cur.remove(cur.size() - 1);
}
return;
}
cur.add(root.val);
pathSum(root.left, sum, results, cur);
pathSum(root.right, sum, results, cur);
cur.remove(cur.size() - 1);
}
}
113,path-sum---https://leetcode.com/problems/path-sum/---NOT BUG FREE
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
CUR:思路不清晰。
思路：如果是叶子节点，并且sum == root.val。那么就可以return true；
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
if (root.left == null && root.right == null) {
if (sum == root.val) {
return true;
}
return false;
}
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
112，minimum depth of binary tree(根结点到叶节点的最小高度)---https://leetcode.com/problems/minimum-depth-of-binary-tree/---BUG FREE
Given a binary tree, find its minimum depth. The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.
CUR：思路错误。
答案和思路：比起高度不同的是，有一边为空的时候是不考虑的，所以，如果为空就返回另一边+1。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public int minDepth(TreeNode root) {
if (root == null) {
return 0;
}
int left = minDepth(root.left);
int right = minDepth(root.right);
if (left == 0) {
return right +1;
}
if (right == 0) {
return left +1;
}
return Math.min(left, right) + 1;
}
}
111,symmetic tree---https://leetcode.com/problems/symmetric-tree/--BUG FREE
101. Symmetric Tree QuestionEditorial Solution My Submissions
Total Accepted: 131244
Total Submissions: 366595
Difficulty: Easy
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/ \
2 2
\ \
3 3
CUR:错误理解了Symmetric。对称指的是从中间一条竖线。
答案和思路。检查的是他的左右结点。他的左儿子的左儿子和他的右儿子的右儿子作比较。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) {
return true;
}
return isSymmetric(root.left, root.right);
}
private static boolean isSymmetric(TreeNode left, TreeNode right) {
if (left == null && right == null) {
return true;
} else if (left == null || right == null) {
return false;
} else {
if (left.val != right.val) {
return false;
} else {
return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
}
}
}
}
110，balanced binary tree---https://leetcode.com/problems/balanced-binary-tree/---BUG FREE
cur：思路不清晰
第二天：注意left == -1 || right == -1也是判断的条件之一。
答案：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
}
return depth(root) != -1;
}
private static int depth(TreeNode root) {
if (root == null) {
return 0;
}
int left = depth(root.left);
int right = depth(root.right);
if (left == -1 || right == -1 || Math.abs(left - right) > 1) {
return -1;
}
return Math.max(left, right) + 1;
}
}
109，求树的高度。---BUG FREE
public int depth(TreeNode root) {
if (root == null) {
;
}
;
}
108，给40亿个数字，看缺少哪个。---NOT BUG FREE
答：对整数做二分，数一下40亿里面比他少的有多少个。看是否有少了。
107---binary search tree find, add, remove
import javax.xml.soap.Node;
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
public class Main {
public static boolean find(int value, TreeNode node) {
while (node != null) {
if (node.val == value) {
return true;
} else if (node.val > value) {
node = node.left;
} else {
node = node.right;
}
}
return false;
}
public static boolean add(int value, TreeNode root) {
if (root == null) {
root = new TreeNode(value);
return true;
}
TreeNode node = root;
while (node != null) {
if (node.val > value) {
if (node.left != null) {
node = node.left;
} else {
node.left = new TreeNode(value);
return true;
}
} else if (node.val < value) {
if (node.right != null) {
node = node.right;
} else {
node.right = new TreeNode(value);
return true;
}
}
}
return false;
}
public static boolean remove(int value, TreeNode root) { // find, then removeNode
if (root == null) {
return false;
}
if (root.val == value) {
root = removeNode(root);
return true;
}
TreeNode node = root;
while (node != null) {
if (node.val > value) {
if (node.left != null && node.left.val != value) {
node = node.left;
} else if (node.left == null) {
return false;
} else {
node.left = removeNode(node.left);
return true;
}
} else if (node.val < value) {
if (node.right != null && node.right.val != value) {
node = node.right;
} else if (node.right != null) {
return false;
} else {
node.right = removeNode(node.right);
return true;
}
} else {
return false;
}
}
return false;
}
public static TreeNode removeNode(TreeNode node) {
if (node.left == null && node.right == null) {
return null;
} else if (node.left == null) {
return node.right;
} else if (node.right == null) {
return node.left;
} else {
node.val = findAndRemove(node);
return node;
}
}
public static int findAndRemove(TreeNode node) {
int result;
if (node.left.right == null) {
result = node.left.val;
node.left = node.left.left;
return result;
}
node = node.left;
while (node.right.right != null) {
node = node.right;
}
result = node.right.val;
node.right = node.right.left;
return result;
}
}
106---binary tree from inorder and postorder traversal---https://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/---NOT BUG FREE
Given inorder and postorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree.
答案和思路：
public class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
int len = inorder.length;
return buildTree(inorder, 0, len - 1, postorder, 0, len - 1);
}
private static TreeNode buildTree(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {
if (inStart > inEnd || postStart > postEnd) {
return null;
}
int val = postorder[postEnd];
TreeNode root = new TreeNode(val);
int k = 0;
for (int i = inStart; i <= inEnd; i++) {
if (val == inorder[i]) {
k = i;
break;
}
}
root.left = buildTree(inorder, inStart, k - 1, postorder, postStart, postStart + k - inStart - 1);
root.right = buildTree(inorder, k + 1, inEnd, postorder, postStart + k - inStart, postEnd - 1);
return root;
}
}
105---binary tree from preorder and inorder traversal---https://leetcode.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/---NOT BUG FREE
Given preorder and inorder traversal of a tree, construct the binary tree. Note: You may assume that duplicates do not exist in the tree. Subscribe to see which companies asked this question
答案和思路：找到根在中序的位置，就确定了左右两边怎么走。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
int len = preorder.length;
return buildTree(preorder, 0, len - 1, inorder, 0, len - 1);
}
private static TreeNode buildTree(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
if (preStart > preEnd || inStart > inEnd) {
return null;
}
int val = preorder[preStart];
TreeNode p = new TreeNode(val);
int k = 0;
for (int i = inStart; i <= inEnd; i++) {
if (val == inorder[i]) {
k = i;
break;
}
}
p.left = buildTree(preorder, preStart + 1, preStart + (k - inStart), inorder, inStart, k - 1);
p.right = buildTree(preorder, preStart + (k - inStart) + 1, preEnd, inorder, k + 1, inEnd);
return p;
}
}
=============================================================bug-free分割线===========================
104---binary tree postOrder(树的后序遍历）---https://leetcode.com/problems/binary-tree-postorder-traversal/---BUG FREE
Given a binary tree, return the postorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [3,2,1].
答案和思路：问题关键就在于它的右边是否已经访问完了。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class TreeNodeWithFlag {
TreeNode node;
boolean flag;
public TreeNodeWithFlag(TreeNode n, boolean value) {
node = n;
flag = value;
}
}
public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList();
Stack<TreeNodeWithFlag> stack = new Stack();
TreeNode curNode = root;
TreeNodeWithFlag newNode;
while (!stack.isEmpty() || curNode != null) {
while (curNode != null) {
newNode = new TreeNodeWithFlag(curNode, false);
stack.push(newNode);
curNode = curNode.left;
}
newNode = stack.pop();
curNode = newNode.node;
if (!newNode.flag) {
newNode.flag = true;
stack.push(newNode);
curNode = curNode.right;
}else {
result.add(curNode.val);
curNode = null;
}
}
return result;
}
}
103---binary tree inOrder(树的中序遍历)---https://leetcode.com/problems/binary-tree-inorder-traversal/---BUG FREE
Given a binary tree, return the inorder traversal of its nodes' values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].
答案和思路：一直把node往stack里面放。一直往左走。知道为null。就开始放值。然后对于那个一node往右走。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList();
if (root == null) {
return result;
}
TreeNode node = root;
Stack<TreeNode> stack = new Stack();
while (!stack.isEmpty() || node != null) {
if (node != null) {
stack.push(node);
node = node.left;
} else {
node = stack.pop();
result.add(node.val);
node = node.right;
}
}
return result;
}
}
102,binary tree preorder（树的前序遍历）---https://leetcode.com/problems/binary-tree-preorder-traversal/---BUG FREE
Given a binary tree, return the preorder traversal of its nodes' values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,2,3].
答案1：先放右再放左，然后pop；
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList();
if (root == null) {
return result;
}
Stack<TreeNode> stack = new Stack();
stack.push(root);
while (!stack.isEmpty()) {
TreeNode node = stack.pop();
result.add(node.val);
if (node.right != null) {
stack.push(node.right);
} if (node.left != null) {
stack.push(node.left);
}
}
return result;
}
}
答案和思路：二叉树的前序遍历。node进来的时候，先把值输出，然后如果右边结点！=null。把结点存入stack，然后往左走。如果左不为空，node = node.left。如果为空，node = stack.pop();
优点：比起两个都pop()进去能够优化到：O（n / 2）；
public class Solution {
public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> result = new ArrayList();
if (root == null) {
return result;
}
Stack<TreeNode> stack = new Stack();
stack.push(root);
TreeNode node = root;
while (!stack.isEmpty()) {
result.add(node.val);
if (node.right != null) {
stack.add(node.right);
}
if (node.left != null) {
node = node.left;
} else {
node = stack.pop();
}
}
return result;
}
}
101, Distinct Subsequences(两个字符串独立子序列)---https://leetcode.com/problems/distinct-subsequences/--- BUG FREE
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not). Here is an example: S = "rabbbit", T = "rabbit" Return 3.
答案和思路：如果第i个字符和第j个字符相等，那么，dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1]; 如果不等，那么，dp[i][j] = dp[i - 1][j]。
public class Solution {
public int numDistinct(String s, String t) {
if (s == null) {
return 0;
}
if (t == null) {
return 1;
}
int lenS = s.length();
int lenT = t.length();
int[][] dp = new int[lenS + 1][lenT + 1];
for (int i = 0; i <= lenS; i++) {
dp[i][0] = 1;
}
for (int i = 1; i <= lenS; i++) {
for (int j = 1; j <= lenT; j++) {
if (s.charAt(i - 1) == t.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
} else {
dp[i][j] = dp[i - 1][j];
}
}
}
return dp[lenS][lenT];
}
}
100，triangle（三角形）---https://leetcode.com/problems/triangle/---BUG FREE
Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
答案和思路：
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
if (null == triangle || triangle.size() == 0) {
return 0;
}
int min = Integer.MAX_VALUE;
int height = triangle.size();
for (int i = height - 2; i >= 0; i--) {
int size = triangle.get(i).size();
for (int j = 0; j < size; j++) {
triangle.get(i).set(j, triangle.get(i).get(j) + Math.min(triangle.get(i + 1).get(j), triangle.get(i + 1).get(j + 1)));
}
}
return triangle.get(0).get(0);
}
}
99，MatrixChain（矩阵乘法）---
Give A0,A1,.....,An-1 n different matrixs Find minimum of multiplication it needs to get the result input: An array P with n+1 numbers A0 = p0*p1, An-1 = Pn-1*Pn
答案和思路：
public static int MatrixChain(int[] p)
{
int n = p.length;
n --;
int[][] m = new int[n][n];
for(int i = 0; i < n; i++)
m[i][i] = 0;
for(int r = 2; r <= n; r++)
{
for(int i = 0; i < n - r + 1; i ++)
{
int j = i + r - 1;
m[i][j] = m[i + 1][j] + p[i] * p[i+1] * p[j + 1];
for(int k = i + 1; k < j; k++)
{
int t = m[i][k] + m[k + 1][j] + p[i] * p[k+1] * p[j+1];
if( t < m[i][j])
m[i][j] = t;
}
}
}
return m[0][n-1];
}
98,Edit Distance(编辑距离)---https://leetcode.com/problems/edit-distance/---BUG FREE
Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.) You have the following 3 operations permitted on a word: a) Insert a character b) Delete a character c) Replace a character
CUR:思路不清晰。忘记初始化了。
答案和思路：dp[i][j]表示的是第i个和第j个之间需要的次数。那么他们是否是相等的字符构成了三种可能
public class Solution {
public int minDistance(String word1, String word2) {
if (word1 == null && word2 == null) {
return 0;
} else if (word1 == null) {
return word2.length();
} else if (word2 == null) {
return word1.length();
}
int len1 = word1.length();
int len2 = word2.length();
int[][] dp = new int[len1 + 1][len2 + 1];
for (int i = 0; i <= len1; i++) {
dp[i][0] = i;
}
for (int j = 0; j <= len2; j++) {
dp[0][j] = j;
}
for (int i = 1; i <= len1; i++) {
for (int j = 1; j <= len2; j++) {
if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j - 1];
} else {
dp[i][j] = Math.min(dp[i - 1][j - 1], Math.min(dp[i][j - 1], dp[i - 1][j])) + 1;
}
}
}
return dp[len1][len2];
}
}
97, longest common subsequence---http://www.lintcode.com/zh-cn/problem/longest-common-subsequence/#---BUG FREE
给出两个字符串，找到最长公共子序列(LCS)，返回LCS的长度。
答案和思路：
public class Solution {
/**
* @param A, B: Two strings.
* @return: The length of longest common subsequence of A and B.
*/
public int longestCommonSubsequence(String A, String B) {
// write your code here
if (A == null || B == null) {
return 0;
}
int lenA = A.length();
int lenB = B.length();
int[][] dp = new int[lenA + 1][lenB + 1];
for (int i = 1; i <= lenA; i++) {
for (int j = 1; j <= lenB; j++) {
if(A.charAt(i - 1) == B.charAt(j - 1)) {
dp[i][j] = Math.max(dp[i - 1][j - 1] + 1, Math.max(dp[i - 1][j], dp[i][j - 1]));
} else {
dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
}
}
}
return dp[lenA][lenB];
}
}
96，longest increasing subsequence---https://leetcode.com/problems/longest-increasing-subsequence/--BUG FREE
Given an unsorted array of integers, find the length of longest increasing subsequence. For example, Given [10, 9, 2, 5, 3, 7, 101, 18], The longest increasing subsequence is [2, 3, 7, 101], therefore the length is 4. Note that there may be more than one LIS combination, it is only necessary for you to return the length. Your algorithm should run in O(n2) complexity.
答案和思路：
public class Solution {
public int lengthOfLIS(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int[] dp = new int[nums.length];
int max = 0;
for (int i = 0; i < nums.length; i++) {
dp[i] = 1;
for (int j = 0; j < i; j++) {
if (nums[j] < nums[i]) {
dp[i] = Math.max(dp[i], dp[j] + 1);
}
}
max = Math.max(dp[i], max);
}
return max;
}
}
95---0-1背包---BUG FREE
0-1 Knapsack Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Each item has its value s1,s2,...,sn. Try to select the items that could put in knapsack and contains most value.
答案和思路：dp[i][j]表示的是前i个物品在容量为j的时候得到的最大值。if第i个物品重量比容量还大，那么就等于前i-1个物品就可以，否则的话，要么加上他的值再加上j-他的重量的那里得到的结果，要么不要他，也就是前i-1个物品。
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Solution {
public static void main(String[] args) {
int[] w = {1, 3, 4, 5};
int[] v = {3, 8, 4, 7};
System.out.println(knapsack(7, w, v));
}
public static int knapsack(int capacity, int[] weights, int[] values) {
if (capacity <= 0 || weights == null || weights.length == 0 || values == null || values.length == 0) {
return 0;
}
int[][] dp = new int[weights.length + 1][capacity + 1];
for (int i = 1; i <= weights.length; i++) {
for (int j = 1; j <= capacity; j++) {
if (j < weights[i - 1]) {
dp[i][j] = dp[i - 1][j];
} else {
dp[i][j] = Math.max(dp[i - 1][j - weights[i - 1]] + values[i - 1], dp[i - 1][j]);
}
}
}
return dp[weights.length][capacity];
}
}
94, Minimum Path Sum(最小路径和)---https://leetcode.com/problems/minimum-path-sum/---BUG FREE
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path.
答案和思路：
public class Solution {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[m][n];
dp[0][0] = grid[0][0];
for (int i = 1; i < m; i++) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (int j = 1; j < n; j++) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[m - 1][n - 1];
}
}
93, Unique Paths II（唯一路径2）---https://leetcode.com/problems/unique-paths-ii/--- Bug Free
Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middle of a 3x3 grid as illustrated below. [ [0,0,0], [0,1,0], [0,0,0] ] The total number of unique paths is 2.
答案和思路：
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (null == obstacleGrid || obstacleGrid.length == 0) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
if (obstacleGrid[0][0] == 1) {
return 0;
}
dp[0][0] = 1;
for (int i = 1; i < m; i++) {
if (obstacleGrid[i][0] == 1) {
break;
}
dp[i][0] = 1;
}
for (int j = 1; j < n; j++) {
if (obstacleGrid[0][j] == 1) {
break;
}
dp[0][j] = 1;
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] != 1) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
}
92, Unique Paths（唯一路径）---https://leetcode.com/problems/unique-paths/--- BUG FREE
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). How many possible unique paths are there?
CUR:注意边界问题。
答案和思路：
public class Solution {
public int uniquePaths(int m, int n) {
if (m <= 0 || n <= 0) {
return 0;
}
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
dp[i][0] = 1;
}
for (int j = 0; j < n; j++) {
dp[0][j] = 1;
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
}
滚动数组优化：
public class Solution {
public int uniquePaths(int m, int n) {
if (m <= 0 || n <= 0) {
return 0;
}
int[][] dp = new int[m][n];
for (int i = 0; i < m; i++) {
dp[i][0] = 1;
}
for (int j = 0; j < n; j++) {
dp[0][j] = 1;
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
}
91,GoValley期末卷子解析---4题耗时2小时10分钟。
Print Path of a Tree Given a Leave
Description:
Given a root of a known binary tree and a key, return the path from the root to this node. We can make sure that there are no duplicates in the tree for the key
Example Tree is like:
Examples:
Key = 30 return [14,9,0,17,30]
Key = 4 return [14,5,8,3,4]
Key = 22 return [14,9,19,22]
API
Java: public List<Integer> PrintPath(Node Root, int key)
C++: public int[] PrintPath(Node* Root, int key)
Java:
Node {
int val;
Node leftChild;
Node rightChild;
}
C++:
Node {
int val;
Node* leftChild;
Node* rightChild;
}
Print Characters from number
Description:
Given a number in string format, return how many character combinations there could be derived from the number. Number to character: 1-A, 2-B, 26-Z. We ensure the string is always valid (it will not contain other character than ‘0’ to ‘9’)
Examples:
Input: 1234
Output: 3,
(These three are the possible results:
<1,2,3,4 ABCD>, <12, 3, 4 LCD>, <1, 23, 4 AWD>
API:
Java: public static int stringCombinationNum(String number);
C++: public int stringCombinationNum(string number);
Division + Cycle
Description:
Given two numbers num1 and num2, return the string format of division result of num1/num2.
If num1 cannot be divided by num2, then put the cycle of the result into brackets.
Examples:
Input: 4, 5
Output: 0.8
Input: 4, 6
Output: 0.(6)
Input: 17, 66
Output: 0.2(57)
API:
Java: public static String divide(int num1, int num2);
C++: public static string divide(int num1, int num2);
Palindrome: Minimum Insertion
Description:
Given a String, return the minimum number of characters that it needs to be inserted to make it become a palindrome. We make sure we only have lowercase English characters (a to z)
Examples:
Input: abc, output: 2 (it can become cbabc or abcba, etc)
Input: ababa, output: 0
Input: abccbac, output: 1 (it will become cabccbac)
API:
Java: public int minimumPalidrome(String word)
C++: public int minimumPalidrome(string word)
期末答案解析：
第一题：
import java.util.ArrayList;
import java.util.List;
class Node {
int val;
Node leftChild;
Node rightChild;
public Node(int x) {
val = x;
}
}
public class Solution {
/**
* 答案和思路：利用树的递归性质，往左往右找，用List存下来路径的每一个点，如果找到了就返回。值得注意的是每一个ArrayList必须新开，不然左右的值都会add到一起。
* 时间复杂度分析：O(n)
* @author ChunyueLi
*/
public static void main(String[] args) {
Node root = new Node(14);
root.leftChild = new Node(5);
root.rightChild = new Node(9);
root.leftChild.leftChild = new Node(8);
root.leftChild.rightChild = new Node(6);
System.out.println(printPath(root, 6));
}
public static List<Integer> printPath(Node root, int key) {
List<Integer> result = new ArrayList();
if (root == null) {
return result;
}
return findPath(root, key, result);
}
private static List<Integer> findPath(Node root, int key, List<Integer> result) {
if (root != null) {
result.add(root.val);
if (root.val == key) {
return new ArrayList(result);
}
} else {
return null;
}
List<Integer> left = findPath(root.leftChild, key, new ArrayList(result));
if (left != null) {
return left;
} else {
return findPath(root.rightChild, key, new ArrayList(result));
}
}
}
第二题：
public class Solution {
public static void main(String[] args) {
String number = "1234";
System.out.println(stringCombinationNum(number));
}
public static int stringCombinationNum(String number) {
/**
* 答案和思路：这道题是一道dp题。解答并不难。dp[i]表示的是前i个字符能够表示多少种合法的字母。那么对于第i个位置，唯一能够影响他的也就是他的前一个和他是否又能够构成一个字母，如果能的话，他就有两种组成方式。
* 一种组成方式是他自己和他的前面的i-1个组成，一种是他和i-1再和他前面的i-2个组成。
* 时间复杂度分析：O（n)
* 滚动指针优化动态规划：%3就可以。空间复杂度从O（n)优化到O（1）
* @author ChunyueLi
*/
int n = number.length();
int[] dp = new int[n+1]; // A table to store results of subproblems
dp[0] = 1;
dp[1] = 1;
for (int i = 2; i <= n; i++)
{
if (number.charAt(i - 1) > '0') {
dp[i] = dp[i-1];
}
if (number.charAt(i - 2) < '2' || (number.charAt(i - 2) == '2' && number.charAt(i-1) < '7') ) {
dp[i] += dp[i-2];
}
}
return dp[n];
}
}
第三题：
import java.util.LinkedHashMap;
public class Solution {
public static void main(String[] args) {
System.out.println(divide2(17, 66));
}
/**
* 这里用LinkedHashMap来看是否已经循环了.LinkedHashMap的好处就是他可以记住插入的顺序位置
*/
public static String divide2(int num1, int num2) {
LinkedHashMap<Integer, Integer> map = new LinkedHashMap();
StringBuffer sb = new StringBuffer();
while (num1 != 0) {
if (map.containsKey(num1)) {
sb = new StringBuffer();
int i = 1;
for (int n : map.keySet()) {
if (n < num2 && sb.length() == 0) {
sb.append("0.");
}
if (n == num1) {
sb.append('(');
sb.append(map.get(n));
} else {
sb.append(map.get(n));
}
i++;
}
sb.append(')');
return sb.toString();
}
if (num1 < num2) {
if (sb.length() == 0) {
sb.append("0.");
} else {
sb.append('.');
}
sb.append(num1 * 10 / num2);
map.put(num1, num1 * 10 / num2);
num1 = num1 * 10 % num2;
} else {
sb.append(num1 / num2);
map.put(num1, num1 / num2);
num1 = num1 % num2;
}
}
return sb.toString();
}
}
第四题：
public class YUEYE {
public static void main(String[] args) {
System.out.println(minimumPalidromw("abc"));
}
/**
* 思路：找到最小的插入位置就核心。
* @author ChunyueLi
*/
public static int minimumPalidromw(String word) {
return helper(word, 0, word.length() - 1);
}
public static int helper(String str, int l, int h)
{
if (l > h) {
return Integer.MAX_VALUE;
}
if (l == h) {
return 0;
}
if (l == h - 1) {
return (str.charAt(l) == str.charAt(h))? 0 : 1;
}
if (str.charAt(l) == str.charAt(h)) {
return helper(str, l + 1, h - 1);
} else {
return Math.min(helper(str, l, h - 1), helper(str, l + 1, h)) + 1;
}
}
}
90，数组次小次大。
用样本中有限的编号来估计总体数量，可以使用“一致最小变异不偏估计量’或‘无偏见最佳估计”理论进行估计假设坦克是按从1到N顺序生产的，通过最小方差无偏估计得出估算：m是（样本最大值）最大序列号，k是观测到的坦克数（样本大小）。一个序列号经过观测后将被排除样本池之外，不再进行观测。方差：但是这个估计是有误差的。真正的坦克总数应该是在估计值左右，所以我们就要用置信区间（ConfidenceInterval）的方式，把真实值左右的变化度体现出来。通过置信区间我们可以结论，有95%信心，真正的坦克数量是出现在这个区间之中[m,m/p^1/k]改估计的优点是用贝叶斯估计改进了传统的统计学理论，比传统估计更加准确，缺点是公式复杂。
import java.util.Arrays;
public class Solution {
static int[] a = {7, 6, 5, 4, 3, 2, 1};
public static void main(String[] args) {
smallBigSort(a); //按照题目的要求进行排序
for (int i = 0; i < a.length; i++) {
System.out.print(a[i] + ","); //排序后打印出来
}
}
public static void smallBigSort(int[] a) {
boolean flag = false;
for (int i = 0; i < a.length - 1; i++) {
if (flag) {
for (int j = a.length - 1; j > i; j--) {
if (a[j] > a[j - 1]) {
int team = a[j];
a[j] = a[j - 1];
a[j - 1] = team;
}
}
} else {
for (int j = a.length - 1; j > i; j--) {
if (a[j] <= a[j - 1]) {
int team = a[j];
a[j] = a[j - 1];
a[j - 1] = team;
}
}
}
flag = !flag;
}
}
}
89,Palindrome Linked List---https://leetcode.com/problems/palindrome-linked-list/--- BUG FREE
Given a singly linked list, determine if it is a palindrome. Follow up: Could you do it in O(n) time and O(1) space?
CUR:判断是否一样的时候，一定用看后面的是否走到了null，前面的不可能走到null的。
第二天：reverse的代码还是不熟悉。
第三天：reverse没写好。
答案和思路：首先找到mid，然后reverse mid后面的部分。注意在比较的时候千万别忘了.next。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public boolean isPalindrome(ListNode head) {
// find the mid, the reverse.
if (head == null || head.next == null) {
return true;
}
// reverse from the mid
ListNode middle = findMiddle(head);
middle.next = reverse(middle.next);
return isSame(head, middle.next);
}
private static ListNode findMiddle(ListNode head) {
ListNode slow = head;
ListNode fast = head;
while (fast.next != null && fast.next.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
private static ListNode reverse(ListNode node) {
ListNode pre = null;
while (node != null) {
ListNode next = node.next;
node.next = pre;
pre = node;
node = next;
}
return pre;
}
private static boolean isSame(ListNode head1, ListNode head2) {
while (head1 != null && head2 != null && head1.val == head2.val) {
head1 = head1.next;
head2 = head2.next;
}
return head2 == null ? true : false;
}
}
88，combinations的Follow Up
Given a collection of numbers C and a number k, return all the combinations of k numbers. Note: There could be duplicates in C, but duplicates are not allowed in the results. Examples: Input: [1, 1, 1], 2 -> Output: [[1, 1]] Input: [2, 3, 4], 2 -> Output: [[2, 3], [2, 4], [3, 4]]
答案和思路：利用sort来排除重复解。
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
public class Solution {
public static void main(String[] args) {
int[] w = {1, 1, 1};
System.out.println(combination(w, 2));
}
public static List<List<Integer>> combination(int[] nums, int k) {
List<List<Integer>> results = new ArrayList();
if (nums == null || nums.length == 0 || k < 0) {
return results;
}
Arrays.sort(nums);
List<Integer> cur = new ArrayList();
combine(nums, k, results, cur, 0);
return results;
}
private static void combine(int[] nums, int k, List<List<Integer>> results, List<Integer> cur, int index) {
if (cur.size() == k) {
results.add(new ArrayList(cur));
return;
}
for (int i = index; i < nums.length; i++) {
if (i != index && nums[i] == nums[i - 1]) {
continue;
}
cur.add(nums[i]);
combine(nums, k, results, cur, i + 1);
cur.remove(cur.size() - 1);
}
}
}
87，combinations---https://leetcode.com/problems/combinations/---BUG FREE
Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. For example, If n = 4 and k = 2, a solution is: [ [2,4], [3,4], [2,3], [1,2], [1,3], [1,4], ]
答案和思路：还依然是利用index坐标，先是add然后递归，然后再remove出来。
public class Solution {
public List<List<Integer>> combine(int n, int k) {
List<List<Integer>> results = new ArrayList();
if (k <= 0 || n <= 0) {
return results;
}
List<Integer> cur = new ArrayList();
combine(results, cur, n, k, 1);
return results;
}
private static void combine(List<List<Integer>> results, List<Integer> cur, int n, int k, int start) {
if (cur.size() == k) {
results.add(new ArrayList(cur));
return;
}
for (int i = start; i <= n; i++) {
cur.add(i);
combine(results, cur, n, k, i + 1);
cur.remove(cur.size() - 1);
}
}
}
86，Given 2 sorted array, find the kth smallest number---BUG FREE
1. Given 2 sorted array, find the kth smallest number. Examples: a. Input: [1, 3, 5], [2, 4, 6], 5 -----> Output: 5 b. Input: [6, 9, 17], [2, 18, 30], 4 -----> Output: 17 Interface: Java: int kthSmall(int[] arr1, int[] arr2, int k); C++: int kthSmall(vector<int>arr1, vector<int>arr2, int k);
O(k)解法：
思路：这里思路利用的是归并的merge部分的思想。
时间复杂度是：0(K).
代码如下：
public int kthSmall(int[] arr1, int[] arr2, int k) {
if (arr1 == null || arr2 == null || arr1.length + arr2.length < k || k <= 0) {
return -1;
}
int arr1Index = 0;
int arr2Index = 0;
int result = -1;
while (arr1Index < arr1.length || arr2Index < arr2.length) {
while (arr1Index < arr1.length &&(arr2Index >= arr2.length || arr1[arr1Index] < arr2[arr2Index])) {
arr1Index++;
// 这里虽然index都是下标，但是由于每一次判断的时候，index都是指向了下一个位置。所以不用再多加1.
if (arr1Index + arr2Index == k) {
return arr1[arr1Index - 1];
}
}
while (arr2Index < arr2.length && (arr1Index >= arr1.length || arr2[arr2Index] < arr1[arr1Index])) {
arr2Index++;
if (arr1Index + arr2Index == k) {
return arr2[arr2Index - 1];
}
}
}
return result;
}
进一步优化的想法：根据m，n，k的范围可以进一步优化，比如，m，n，k都非常大，k也远远大于m+n的一半，那么从后往前找能够优化。
Olog(k)的解法：
int kthSmall(int[] arr1, int[] arr2, int k){
return findKthSmallest(arr1, arr2, 0, 0, arr1.length-1, arr2.length-1, k);
}
int findKthSmallest(int a[], int b[], int startA, int startB, int endA, int endB, int k){
if(endA-startA > endB - startB) {
return findKthSmallest(b,a,startB,startA,endB,endA,k);
}
if(endA<startA) return b[startB+k-1];
if(k == 1) return Math.min(a[startA], b[startB]);
int pA = Math.min(endA-startA + 1, k/2);
int pB = k - pA;
if(a[startA + pA - 1] < b[startB + pB - 1]) {
return findKthSmallest(a,b,startA + pA, startB, endA, startB + pB, k - pA);
}
else if(a[startA + pA - 1] > b[startB + pB - 1]) {
return findKthSmallest(a,b,startA, startB + pB, startA + pA, endB, k - pB);
}
else {
return a[startA + pA - 1];
85,generate parentheses---https://leetcode.com/problems/generate-parentheses/---BUG FREE
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses. For example, given n = 3, a solution set is: [ "((()))", "(()())", "(())()", "()(())", "()()()" ]
CUR:注意left， right 表示的是还剩下几个需要放进去。
第二天：
答案和思路：控制好left，right。
public class Solution {
public List<String> generateParenthesis(int n) {
List<String> results = new ArrayList();
generateParenthesis(results, "", n, n);
return results;
}
private static void generateParenthesis(List<String> results, String prefix, int left, int right) {
if (left == 0 && right == 0) {
results.add(prefix);
return;
}
if (left > 0) {
generateParenthesis(results, prefix + "(", left - 1, right);
}
if (right > 0 && left < right) {
generateParenthesis(results, prefix + ")", left, right - 1);
}
}
}
84,merge intervals---https://leetcode.com/problems/merge-intervals/---BUG FREE
Given a collection of intervals, merge all overlapping intervals. For example, Given [1,3],[2,6],[8,10],[15,18], return [1,6],[8,10],[15,18].
CUR:重排序，new Comparator<Interval>() { public int compare(Interval a, Interval b) { return }};
第二天：忘记了最后的那个值也要加进去。然后还有就是end只要大于等于start就要合并，而不需要严格大于。
第三次，忘记了最后那个值也要加进去。
答案和思路：按照start排序，然后对每一个进行处理，如果他的end比他的下一个的start大，那么就把他的end换成两个钟比较大的那个，如果没有交集，那么就存入结果。
public class Solution {
public List<Interval> merge(List<Interval> intervals) {
List<Interval> result = new ArrayList();
if (intervals == null || intervals.size() == 0) {
return result;
}
Collections.sort(intervals, new Comparator<Interval>() {
public int compare(Interval a, Interval b) {
return a.start - b.start;
}
});
Interval pre = intervals.get(0);
for (int i = 1; i < intervals.size(); i++) {
if (intervals.get(i).start <= pre.end) {
pre.end = Math.max(intervals.get(i).end, pre.end);
} else {
result.add(new Interval(pre.start, pre.end));
pre = intervals.get(i);
}
}
result.add(pre);
return result;
}
}
84, maximum subarray---https://leetcode.com/problems/maximum-subarray/--BUG FREE
Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] has the largest sum = 6.
答案和思路：
public class Solution {
public int maxSubArray(int[] nums) {
int[] dp = new int[nums.length];
int max = nums[0];
dp[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
dp[i] = Math.max(nums[i], dp[i - 1] + nums[i]);
max = Math.max(max, dp[i]);
}
return max;
}
}
83,minimum size subarray sum---https://leetcode.com/problems/minimum-size-subarray-sum/---BUG FREE
Given an array of n positive integers and a positive integer s, find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead. For example, given the array [2,3,1,2,4,3] and s = 7, the subarray [4,3] has the minimal length under the problem constraint. click to show more practice. Credits:
CUR:注意i要走到最后的。然后题目要求的是sum >= s.
第二天：还是老错误，题目要求的是sum>=s而不是sum == s。
答案和思路：不后退的双指针。
public class Solution {
public int minSubArrayLen(int s, int[] nums) {
if (s == 0 || nums == null || nums.length == 0) {
return 0;
}
int sum = 0;
int end = 0;
int result = Integer.MAX_VALUE;
for (int i = 0; i < nums.length; i++) {
while (end < nums.length && sum < s) {
sum += nums[end++];
}
if (sum >= s) {
result = Math.min(result, end - i);
}
sum -= nums[i];
}
return result == Integer.MAX_VALUE ? 0 : result;
}
}
82，---N-Queens---https://leetcode.com/problems/n-queens-ii/--BUG FREE
Follow up for N-Queens problem. Now, instead outputting board configurations, return the total number of distinct solutions.
答案和思路：
public class Solution {
public int totalNQueens(int n) {
if (n <= 0) {
return 0;
}
int[] count = new int[1];
int[] board = new int[n];
placeNQueens(count, board, 0);
return count[0];
}
private static void placeNQueens(int[] count, int[] board, int row) {
if (row == board.length) {
count[0]++;
return;
}
for (int col = 0; col < board.length; col++) {
if (isLegal(board, row, col)) {
board[row] = col;
placeNQueens(count, board, row + 1);
}
}
}
private static boolean isLegal(int[] board, int row, int col) {
for (int i = row - 1; i >= 0; i--) {
if (board[i] == col || Math.abs(col - board[i]) == row - i) {
return false;
}
}
return true;
}
}
81,---N-Queens---https://leetcode.com/problems/n-queens/--BUG FREE
Given an integer n, return all distinct solutions to the n-queens puzzle. Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively. For example, There exist two distinct solutions to the 4-queens puzzle: [ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ]
答案和分析：
public class Solution {
public List<List<String>> solveNQueens(int n) {
List<List<String>> results = new ArrayList();
if (n <= 0) {
return results;
}
int[] board = new int[n];
placeNQueens(results, board, 0);
return results;
}
private static void placeNQueens(List<List<String>> results, int[] board, int row) {
if (row == board.length) {
save(results, board);
return;
}
for (int col = 0; col < board.length; col++) {
if (isLegal(board, row, col)) {
board[row] = col;
placeNQueens(results, board, row + 1);
}
}
}
private static boolean isLegal(int[] board, int row, int col) {
for (int i = row - 1; i >= 0; i--) {
if (board[i] == col || Math.abs(col - board[i]) == row - i) {
return false;
}
}
return true;
}
private static void save(List<List<String>> results, int[] board) {
List<String> cur = new ArrayList();
for (int i = 0; i < board.length; i++) {
StringBuffer sb = new StringBuffer();
for (int j = 0; j < board.length; j++) {
if (j == board[i]) {
sb.append('Q');
} else {
sb.append('.');
}
}
cur.add(new String(sb));
}
results.add(cur);
}
}
80,---combination sum II---https://leetcode.com/problems/combination-sum-ii/---BUG FREE
Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Each number in C may only be used once in the combination. Note: All numbers (including target) will be positive integers. The solution set must not contain duplicate combinations. For example, given candidate set [10, 1, 2, 7, 6, 1, 5] and target 8, A solution set is: [ [1, 7], [1, 2, 5], [2, 6], [1, 1, 6] ]
CUR:注意，第一需要排序。因为只有排序才能够利用相邻跳过法。一定要排序，因为不重复解。[1 , 7] [7, 1]不排序就分不开了。
第二天：注意要排序从能去重复解。然后注意while去重的位置。
public class Solution {
public List<List<Integer>> combinationSum2(int[] candidates, int target) {
List<List<Integer>> results = new ArrayList();
if (candidates == null || candidates.length == 0) {
return results;
}
List<Integer> cur = new ArrayList();
Arrays.sort(candidates);
combine(candidates, target, 0, results, cur);
return results;
}
private static void combine(int[] candidates, int target, int index, List<List<Integer>> results, List<Integer> cur) {
if (target == 0) {
results.add(new ArrayList(cur));
return;
}
if (target < 0 || index == candidates.length) {
return;
}
for (int i = index; i < candidates.length; i++) {
cur.add(candidates[i]);
combine(candidates, target - candidates[i], i + 1, results, cur);
cur.remove(cur.size() - 1);
while (i < candidates.length - 1 && candidates[i] == candidates[i + 1]) {
i++;
}
}
}
}
79,---combination sum---https://leetcode.com/problems/combination-sum/--- BUG FREE
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. The same repeated number may be chosen from C unlimited number of times. Note: All numbers (including target) will be positive integers. The solution set must not contain duplicate combinations. For example, given candidate set [2, 3, 6, 7] and target 7, A solution set is: [ [7], [2, 2, 3] ]
CUR：注意添加cur的时候一定要新new ArrayList。另外注意一下如果target小于0就要return。因为是可以重复使用，如果不保证是整数的话会一直往下走。
第二天：忘记return。
第三天:target 忘记-candidates[i]。
第四天：new ArrayList(cur)。
答案和思路：注意，既然题目要求不能有重复的组合，那么还是要排序的。
public class Solution {
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> results = new ArrayList();
if (candidates == null || candidates.length == 0) {
return results;
}
List<Integer> cur = new ArrayList();
Arrays.sort(candidates);
combine(candidates, target, 0, results, cur);
return results;
}
private static void combine(int[] candidates, int target, int index, List<List<Integer>> results, List<Integer> cur) {
if (target == 0) {
results.add(new ArrayList(cur));
return;
}
if (target < 0 || index == candidates.length) {
return;
}
for (int i = index; i < candidates.length; i++) {
cur.add(candidates[i]);
combine(candidates, target - candidates[i], i, results, cur);
cur.remove(cur.size() - 1);
while (i < candidates.length - 1 && candidates[i] == candidates[i + 1]) {
i++;
}
}
}
}
78,letter combinations of a phone number---https://leetcode.com/problems/letter-combinations-of-a-phone-number/---BUG FREE
17. Letter Combinations of a Phone Number QuestionEditorial Solution My Submissions Total Accepted: 96686 Total Submissions: 316654 Difficulty: Medium Given a digit string, return all possible letter combinations that the number could represent. A mapping of digit to letters (just like on the telephone buttons) is given below.
CUR: 注意string的length()；
第二天：注意建立成Map<Character, new String[]>这样后面就不用再不断地new String了。这里面的new是错的。
答案和分析：建立map。思路就是把每一个个数字对应的字母都添加到之前已经存在的结果的每一个后面。所以，在开始之前需要add("")作为初始化。
public class Solution {
public static List<String> letterCombinations(String digits) {
List<String> result = new ArrayList();
if (digits == null || digits.length() == 0) {
return result;
}
Map<Character, String[]> map = new HashMap<Character, String[]>();
// map.put('0', new String[] {});
// map.put('1', new String[] {});
map.put('2', new String[] { "a", "b", "c" });
map.put('3', new String[] { "d", "e", "f" });
map.put('4', new String[] { "g", "h", "i" });
map.put('5', new String[] { "j", "k", "l" });
map.put('6', new String[] { "m", "n", "o" });
map.put('7', new String[] { "p", "q", "r", "s" });
map.put('8', new String[] { "t", "u", "v"});
map.put('9', new String[] { "w", "x", "y", "z" });
result.add("");
for (int i = 0; i < digits.length(); i++) {
String[] cur = map.get(digits.charAt(i));
List<String> temp = new ArrayList();
for (String str : result) {
for (int j = 0; j < cur.length; j++) {
String strTemp = new String(str);
strTemp += cur[j];
temp.add(strTemp);
}
}
result = new ArrayList(temp);
}
return result;
}
}
77,knapsackIII-follow up---BUG FREE
0-1 Knapsack (Required) Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Try to put items into the pack as many as possible, print out all the items that can get the largest weight. Each item can only get picked once.
答案：先找到最大，然后再看能够成最大的组合有哪些？
public static ArrayList<ArrayList<Integer>> knapsackPrint(int s, int[] weights) {
// get the largest weight
int max = knapsackMax(s, weights, 0);
ArrayList<ArrayList<Integer>> results = new ArrayList<>();
results = combinationSum(weights, max);
return results;
}
public static int knapsackMax(int s, int[] weights, int index) {
if (s == 0 || index == weights.length) {
return 0;
}
if (weights[index] > s) {
return knapsackMax(s, weights, index + 1);
}
return Math.max(knapsackMax(s, weights, index + 1),
weights[index] + knapsackMax(s - weights[index], weights, index + 1));
}
public static ArrayList<ArrayList<Integer>> combinationSum(int[] candidates, int target) {
Arrays.sort(candidates);
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
combine(candidates, 0, target, results, new ArrayList<Integer>());
return results;
}
public static void combine(int[] candidates, int index, int target,
ArrayList<ArrayList<Integer>> results,
ArrayList<Integer> cur) {
if (target < 0) {
return;
}
if (target == 0) {
results.add(new ArrayList<Integer>(cur));
return;
}
for (int i = index; i < candidates.length; i++) {
cur.add(candidates[i]);
combine(candidates, i + 1, target-candidates[i], results, cur);
cur.remove(cur.size()-1);
while (i < candidates.length - 1 && candidates[i] == candidates[i + 1]) {
i++;
}
}
return;
}
77，subsets II---https://leetcode.com/problems/subsets-ii/--NOT BUG FREE
Given a collection of integers that might contain duplicates, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If nums = [1,2,2], a solution is: [ [2], [1], [1,2,2], [2,2], [1,2], [] ]
答案和思路：问题就在于如何去重。排序之后，在添加的时候，相同的只添加一次。所以就是在添加的时候去重。
public class Solution {
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> results = new ArrayList();
if (nums == null) {
return results;
}
List<Integer> cur = new ArrayList();
Arrays.sort(nums);
helper(nums, results, cur, 0);
return results;
}
private static void helper(int[] nums, List<List<Integer>> results, List<Integer> cur, int index) {
results.add(new ArrayList(cur));
for (int i = index; i < nums.length; i++) {
if (i != index && nums[i] == nums[i - 1]) {
continue;
}
cur.add(nums[i]);
helper(nums, results, cur, i + 1);
cur.remove(cur.size() - 1);
}
}
}
76,subsets---https://leetcode.com/problems/subsets/---NOT BUG FREE
78. Subsets QuestionEditorial Solution My Submissions Total Accepted: 113481 Total Submissions: 334182 Difficulty: Medium Given a set of distinct integers, nums, return all possible subsets. Note: The solution set must not contain duplicate subsets. For example, If nums = [1,2,3], a solution is: [ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
答案：用递归来写能更好的理解递归。思路是：每次进去的时候就把当前的cur存下来。那么他后面可以加上后面的任何一个就能够构成新的subset。
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> result = new ArrayList();
if (nums == null) {
return result;
}
List<Integer> cur = new ArrayList();
helper(nums, result, cur, 0);
return result;
}
private static void helper(int[] nums, List<List<Integer>> result, List<Integer> cur, int index) {
result.add(new ArrayList(cur));
for (int i = index; i < nums.length; i++) {
cur.add(nums[i]);
helper(nums, result, cur, i + 1);
cur.remove(cur.size() - 1);
}
}
}
public class Solution {
public List<List<Integer>> subsets(int[] nums) {
List<List<Integer>> results = new ArrayList();
List<Integer> cur = new ArrayList();
results.add(cur);
if (nums == null || nums.length == 0) {
return results;
}
Arrays.sort(nums);
for (int i = 0; i < nums.length; i++) {
List<List<Integer>> temp = new ArrayList(results);
for (List<Integer> now : results) {
List<Integer> nowTemp = new ArrayList(now);
nowTemp.add(nums[i]);
temp.add(nowTemp);
}
results = new ArrayList(temp);
}
return results;
}
}
75,permutationsII---https://leetcode.com/problems/permutations-ii/---BUG FREE
47. Permutations II QuestionEditorial Solution My Submissions Total Accepted: 85003 Total Submissions: 287703 Difficulty: Medium Given a collection of numbers that might contain duplicates, return all possible unique permutations. For example, [1,1,2] have the following unique permutations: [ [1,1,2], [1,2,1], [2,1,1] ]
答案：
public class Solution {
public static List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> results = new ArrayList();
List<Integer> listNow = new ArrayList();
results.add(listNow);
if (nums == null || nums.length == 0) {
return results;
}
for (int i = 0; i < nums.length; ++i) {
List<List<Integer>> temp = new ArrayList();
for (List<Integer> list : results) {
int size = list.size();
for (int j = 0; j <= size; ++j) {
List<Integer> cur = new ArrayList(list);
cur.add(j, nums[i]);
if (!temp.contains(cur)) {
temp.add(cur);
}
}
}
results = new ArrayList(temp);
}
return results;
}
}
74，permutations---https://leetcode.com/problems/permutations/---BUG FREE
Given a collection of distinct numbers, return all possible permutations. For example, [1,2,3] have the following permutations: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ]
答案：
public class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> results = new ArrayList();
List<Integer> cur = new ArrayList();
results.add(cur);
if (nums == null || nums.length == 0) {
return results;
}
for (int i = 0; i < nums.length; i++) {
List<List<Integer>> temp = new ArrayList();
for (List<Integer> now : results) {
for (int j = 0; j <= now.size(); j++) {
List<Integer> nowTemp = new ArrayList(now);
nowTemp.add(j, nums[i]);
temp.add(nowTemp);
}
}
results = new ArrayList(temp);
}
return results;
}
}
递归：
public class Solution {
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> results = new ArrayList();
if (nums == null || nums.length == 0) {
return results;
}
List<Integer> cur = new ArrayList();
helper(nums, results, cur);
return results;
}
public static void helper(int[] nums, List<List<Integer>> results, List<Integer> cur) {
if (cur.size() == nums.length) {
results.add(new ArrayList(cur));
}
for (int i = 0; i < nums.length; i++) {
if (cur.contains(nums[i])) {
continue;
}
cur.add(nums[i]);
helper(nums, results, cur);
cur.remove(cur.size() - 1);
}
}
}
73，knapsackIII---不可以重复使用同一个元素。---BUG FREE
Knapsack III Given a collection of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T. Candidate numbers may contain duplicate. Each number in C may only be used once in the combination.
答案和思路：如果说
import java.util.ArrayList;
import java.util.List;
public class Solution {
public static void main(String[] args) {
int[] w = {2, 3, 6, 7};
System.out.println(knapsackIII(w, 7));
}
public static List<List<Integer>> knapsackIII(int[] w, int sum) {
List<List<Integer>> results = new ArrayList();
List<Integer> cur = new ArrayList();
combine(w, sum, 0, results, cur);
return results;
}
public static void combine(int[] w, int sum, int index, List<List<Integer>> results, List<Integer> cur) {
if (sum == 0) {
results.add(new ArrayList(cur));
return;
}
if (sum < 0 || index == w.length) {
return;
}
for (int i = index; i < w.length; i++) {
cur.add(w[i]);
combine(w, sum - w[i], i + 1, results, cur);
cur.remove(cur.size() - 1);
while (i < w.length - 1 && w[i] == w[i + 1]) {
i++;
}
}
}
}
72，knapsackIII---给一个目标值，找出所有组合，可以重复使用元素。--- BUG FREE
Given a set of candidate numbers (C) and a target number (T), find all unique combinations in C where the candidate numbers sums to T.
答案和思路：一定要注意，每次存cur的时候必须要新建一个ArrayList。
import java.util.ArrayList;
import java.util.List;
public class Solution {
public static void main(String[] args) {
int[] w = {2, 3, 6, 7};
System.out.println(knapsackIII(w, 7));
}
public static List<List<Integer>> knapsackIII(int[] w, int sum) {
List<List<Integer>> results = new ArrayList();
List<Integer> cur = new ArrayList();
combine(w, sum, 0, results, cur);
return results;
}
public static void combine(int[] w, int sum, int index, List<List<Integer>> results, List<Integer> cur) {
if (sum == 0) {
results.add(new ArrayList(cur));
return;
}
if (sum < 0 || index == w.length) {
return;
}
for (int i = index; i < w.length; i++) {
cur.add(w[i]);
combine(w, sum - w[i], i, results, cur);
cur.remove(cur.size() - 1);
}
}
}
71,0-1knapsack II---看最多能够装多少个---BUG FREE
Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Try to put items into the pack as many as possible, return the largest weight we can get in the knapsack.
答案：要么选，要么不选。
public static int knapsackII(int[] w, int sum, int index) {
if (sum == 0 || index == w.length) {
return 0;
}
if (w[index] > sum) {
return knapsackII(w, sum, index + 1);
}
return Math.max(knapsackII(w, sum, index + 1), w[index] + knapsackII(w, sum - w[index], index + 1));
}
70,0-1knapsack---看是否能装满；(其实就是给一个目标值，找一个能组成他的。)----BUG FREE
0-1 Knapsack Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Return whether we can pick specific items so that their total weight s.
答案：注意，一定要先判断sum是否==0.比如走到了最后一个。已经是0了。
public class Solution {
public static void main(String[] args) {
int[] w = {14, 8, 7, 5, 3};
System.out.println(knapsack(w, 10, 0));
}
public static boolean knapsack(int[] w, int sum, int index) {
if (sum == 0) {
return true;
}
if (index == w.length || sum < 0) {
return false;
}
return knapsack(w, sum - w[index], index + 1) || knapsack(w, sum, index + 1);
}
}
69，N Queens---打印所有摆放方式-- BUG FREE
Cur:思路不清晰。就是利用一个n维的数组，从0行开始放。每一行一列一列的试。然后在此基础上往下一行放，又是一列一列的放。
第二天：注意问题所在。结果指的是：最外层list存的是所有的摆放结果。第二层list存的是一个结果。那么对于这个list里面，有n个string，每一个代表每一行怎么放。
答案和思路：board[i] = j : in the row, put the queen on the j column.
import java.util.ArrayList;
import java.util.List;
public class Solution {
public static void main(String[] args) {
System.out.println(solveNQueens(8).size());
}
public static List<List<String>> solveNQueens(int n) {
List<List<String>> results = new ArrayList();
int[] board = new int[n];
placeNQueens(board, 0, results);
return results;
}
public static void placeNQueens(int[] board, int row, List<List<String>> results) {
if (row == board.length) {
save(board, results);
return;
}
for (int col = 0; col < board.length; col++) {
if (isLegal(row, col, board)) {
board[row] = col;
placeNQueens(board, row + 1, results);
}
}
}
public static boolean isLegal(int row, int col, int[] board) {
for (int i = row - 1; i >= 0; i--) {
if (board[i] == col || Math.abs(board[i] - col) == row - i) {
return false;
}
}
return true;
}
public static void save(int[] board, List<List<String>> results) {
List<String> curRow = new ArrayList();
for (int i = 0; i < board.length; i++) {
StringBuffer oneLine = new StringBuffer();
for (int j = 0; j < board.length; j++) {
if (board[i] == j) {
oneLine.append("Q");
} else {
oneLine.append(".");
}
}
curRow.add(oneLine.toString());
}
results.add(new ArrayList(curRow));
}
}
68,Maze迷宫---BUG FREE
Given a maze and a start point and a target point, print out the path to reach the target.
答案和分析：
public class Solution {
public static boolean solveMaze(char[][] maze, int startX, int startY, int targetX, int targetY, String path) {
if (startX < 0 || startX >= maze.length || startY < 0 || startY >= maze[0].length) {
return false;
}
if (startX == targetX && startY == targetY) {
System.out.println(path);
return true;
}
maze[startX][startY] = 'X';
int[] dx = {1, 0, -1, 0};
int[] dy = {0, 1, 0, -1};
char[] direction = {'D', 'R', 'U', 'L'};
for (int i = 0; i < 4; i++) {
String newPath = path + direction[i] + " ";
if (solveMaze(maze, startX + dx[i], startY + dy[i], targetX, targetY, newPath)) {
return true;
}
}
return false;
}
}
67,knapsack背包---BUG FREE
0-1 Knapsack Given a knapsack which can hold s pounds of items, and a set of items with weight w1, w2, ... wn. Return whether we can pick specific items so that their total weight s. Example Input: s = 20; w = [14, 8, 7, 5, 3]; Example Output: True;
答案和分析：分为选和没有选。
这里都是重量，也就是说都是正数。
public class Solution {
public static void main(String[] args) {
int s = 20;
int[] w = {14, 8, 12, 5, 3};
System.out.println(knapsack(s, w, 0));
}
public static boolean knapsack(int sum, int[] weights, int index) {
if (sum == 0) {
return true;
}
if (sum < 0 || index >= weights.length) {
return false;
}
return knapsack(sum - weights[index], weights, index + 1) || knapsack(sum, weights, index + 1);
}
}
如果不完全都是正数的话把sum < 0 这个判断去掉。
public class Solution {
public static void main(String[] args) {
int[] w = {-14, -6, 7, 5, -3};
System.out.println(knapsack(w, -17, 0));
}
public static boolean knapsack(int[] w, int sum, int index) {
if (sum == 0) {
return true;
}
if (index == w.length) {
return false;
}
return knapsack(w, sum - w[index], index + 1) || knapsack(w, sum, index + 1);
}
}
66，Hanoi----BUG FREE
(1)算步数(2^n - 1)
public class Solution {
public int hanoi(int n) {
if (n == 1) {
return 1;
}
return hanoi(n - 1) + 1 + hanoi(n - 1);
}
}
(2)打印所有步骤：
public class Solution {
public static void main(String[] args) {
hanoi(3, 'A', 'B', 'C');
}
public static void hanoi(int n, char A, char B, char C) {
if (n == 1) {
System.out.println("Move " + A + " to " + C);
return;
}
hanoi(n - 1, A, C, B);
System.out.println("Move " + A + " to " + C);
hanoi(n - 1, B, A, C);
}
}
答案和思路：注意汉诺塔的关键点：
move n-1 disk from A to B;
move the nth dist from A to C;
move the n - 1 disk from B to C;
65，---climb（爬楼问题）---- BUG FREE
There is a building with n stairs, one person can climb 1 or 2 stairs at one time. Print all the possible ways to climb to the top.
CUR:注意选择走两步的时候，1 1 这个已经包含在1那里了。
答案和思路：
public class Solution {
public static void main(String[] args) {
climb(3,"");
}
public static void climb(int n, String preWay) {
if (n == 1) {
System.out.println(preWay + " 1");
return;
}
if (n == 2) {
System.out.println(preWay + " 2");
System.out.println(preWay + " 1 1");
return;
}
climb(n - 1, preWay + " 1");
climb(n - 2, preWay + " 2");
}
}
64,container with most water(装水最大容器)---https://leetcode.com/problems/container-with-most-water/---BUG FREE
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
答案和思路：利用双指针，谁小谁往里走，因为这已经是它能够成最大值的时候了。
public class Solution {
public int maxArea(int[] height) {
if (null == height || height.length == 0) {
return 0;
}
int result = 0;
int left = 0;
int right = height.length - 1;
while (left < right) {
result = Math.max(result, (right - left) * Math.min(height[left], height[right]));
if (height[left] < height[right]) {
left++;
} else {
right--;
}
}
return result;
}
}
63, trapping rain water(灌水)---https://leetcode.com/problems/trapping-rain-water/---BUG FREE
Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.
答案和思路：利用双指针，从外往里走，那么较低的就是预先处理，因为高的这一边决定了他可以和他矮的构成灌水。
public class Solution {
public int trap(int[] height) {
if (null == height || height.length == 1) {
return 0;
}
int left = 0;
int right = height.length - 1;
int result = 0;
while (left < right) {
if (height[left] < height[right]) {
int smaller = height[left];
while (left < right && height[left] <= smaller) {
result += smaller - height[left];
left++;
}
} else {
int smaller = height[right];
while (left < right && height[right] <= smaller) {
result += smaller - height[right];
right--;
}
}
}
return result;
}
}
62, implement stack using queues---https://leetcode.com/problems/implement-stack-using-queues/--- BUG FREE
Cur:注意queue的poll使用。
Implement the following operations of a stack using queues. push(x) -- Push element x onto stack. pop() -- Removes the element on top of the stack. top() -- Get the top element. empty() -- Return whether the stack is empty. Notes: You must use only standard operations of a queue -- which means only push to back, peek/pop from front, size, and is empty operations are valid. Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue. You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).
答案和思路：主要操作在push的时候，保持一个是空的，每次都把他先放入空的，然后把另一个队列的其他值叶放入这个队列，这样就保证了他最后来却最先走。
class MyStack {
// Push element x onto stack.
Queue<Integer> queue1;
Queue<Integer> queue2;
public MyStack() {
queue1 = new LinkedList();
queue2 = new LinkedList();
}
public void push(int x) {
if (queue1.isEmpty()) {
queue1.add(x);
while (!queue2.isEmpty()) {
queue1.add(queue2.poll());
}
} else {
queue2.add(x);
while (!queue1.isEmpty()) {
queue2.add(queue1.poll());
}
}
}
// Removes the element on top of the stack.
public void pop() {
if (!queue1.isEmpty()) {
queue1.poll();
} else {
queue2.poll();
}
}
// Get the top element.
public int top() {
if (queue1.isEmpty()) {
return queue2.peek();
} else {
return queue1.peek();
}
}
// Return whether the stack is empty.
public boolean empty() {
return queue1.isEmpty() && queue2.isEmpty();
}
}
61,implement Queue using stacks---https://leetcode.com/problems/implement-queue-using-stacks/---BUG FREE
Implement the following operations of a queue using stacks. push(x) -- Push element x to the back of queue. pop() -- Removes the element from in front of queue. peek() -- Get the front element. empty() -- Return whether the queue is empty. Notes: You must use only standard operations of a stack -- which means only push to top, peek/pop from top, size, and is empty operations are valid. Depending on your language, stack may not be supported natively. You may simulate a stack by using a list or deque (double-ended queue), as long as you use only standard operations of a stack. You may assume that all operations are valid (for example, no pop or peek operations will be called on an empty queue).
答案和分析：two stack
class MyQueue {
// Push element x to the back of queue.
Stack<Integer> stack1;
Stack<Integer> stack2;
public MyQueue() {
stack1 = new Stack();
stack2 = new Stack();
}
public void push(int x) {
stack1.push(x);
}
// Removes the element from in front of queue.
public void pop() {
if (stack2.isEmpty()) {
while (!stack1.isEmpty()) {
stack2.push(stack1.pop());
}
}
stack2.pop();
}
// Get the front element.
public int peek() {
if (stack2.isEmpty()) {
while (!stack1.isEmpty()) {
stack2.push(stack1.pop());
}
}
return stack2.peek();
}
// Return whether the queue is empty.
public boolean empty() {
if (stack1.isEmpty() && stack2.isEmpty()) {
return true;
}
return false;
}
}
60，simplify path---https://leetcode.com/problems/simplify-path/---BUG FREE
Given an absolute path for a file (Unix-style), simplify it. For example, path = "/home/", => "/home" path = "/a/./b/../../c/", => "/c"
CUR: 注意string的判断相等必须用equals。还要注意输出还要加上/。还有stirng的length();
CUR:注意，要剔除的是""而不是“ ”.
答案和分析：
import java.util.Stack;
public class Solution {
public String simplifyPath(String path) {
Stack<String> stack = new Stack<String>();
String[] strArray = path.split("/");
for (int i = 0; i < strArray.length; i++) {
String str = strArray[i];
if (str.equals("") || str.equals(".")) {
continue;
} else if (str.equals("..")) {
if (!stack.isEmpty()) {
stack.pop();
}
} else {
stack.push(str);
}
}
if (stack.isEmpty()) {
return "/";
}
StringBuilder result = new StringBuilder();
while (!stack.isEmpty()) {
String temp = "/" + stack.pop();
result.insert(0, temp);
}
return result.toString();
}
}
59，largest rectangle in histogram---https://leetcode.com/problems/largest-rectangle-in-histogram/---BUG FREE
import java.util.Stack;
public class Solution {
public int largestRectangleArea(int[] heights) {
if (heights == null || heights.length == 0) {
return 0;
}
int area = 0;
Stack<Integer> stack = new Stack();
for (int i = 0; i < heights.length; i++) {
if (stack.isEmpty() || heights[stack.peek()] < heights[i]) {
stack.push(i);
} else {
int start = stack.pop();
int width = stack.isEmpty() ? i : i - 1 - stack.peek();
area = Math.max(area, heights[start] * width);
i--;
}
}
while (!stack.isEmpty()) {
int start = stack.pop();
int width = stack.isEmpty() ? heights.length : heights.length - stack.peek() - 1;
area = Math.max(area, heights[start] * width);
}
return area;
}
}
答案和分析：用单调栈找他的右边界。
第二天所犯错误：注意存的是他的下标。而不是他的值本身。到需要他pop的时候，这个比他小。他要用的是i - 他的前一个位置+1.
public class Solution {
public int largestRectangleArea(int[] heights) {
if (null == heights || heights.length == 0) {
return 0;
}
int area = 0;
Stack<Integer> stack = new Stack();
for (int i = 0; i < heights.length; i++) {
if (stack.isEmpty() || heights[stack.peek()] < heights[i]) {
stack.push(i);
} else {
int index = stack.pop(); //现在处理的是他。
int width = stack.isEmpty() ? i : i - (stack.peek() + 1); // 他的宽度是在从i这个位置减去他的前一个的下一个位置
area = Math.max(area, heights[index] * width);
i--;
}
}
while (!stack.isEmpty()) {
int index = stack.pop();
int width = stack.isEmpty() ? heights.length : heights.length - (stack.peek() + 1);
area = Math.max(area, heights[index] * width);
}
return area;
}
}
58, bad hair day (about cow, right)
All the cows are facing right, and they can only see other cows in front of it and shorter than itself. It can be blocked by other taller cows. Example: 6 2 3 1 7 4 Output: 5
答案和解析：利用单调栈。因为每一个后面的它能够被几个人看见，所以他入栈的时候他前面的都比他大。那么，他出栈的时候能看见他的就是他pop之后的size。
public static int seeCows(int[] cow) {
int length = cow.length;
int totalNum = 0;
Stack<Integer> cowStack = new Stack<Integer>();
for(int i = 0; i < length; i ++){
if(cowStack.size() == 0 || cowStack.peek() > cow[i]) {
cowStack.push(cow[i]);
}
else {
while(cowStack.size() > 0 && cowStack.peek() <= cow[i]) {
cowStack.pop();
totalNum += cowStack.size();
}
cowStack.push(cow[i]);
}
}
while(cowStack.size() > 0) {
cowStack.pop();
totalNum += cowStack.size();
}
return totalNum;
}
57,min stack---https://leetcode.com/problems/min-stack/---BUG FREE
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. push(x) -- Push element x onto stack. pop() -- Removes the element on top of the stack. top() -- Get the top element. getMin() -- Retrieve the minimum element in the stack.
答案和思路：不需要建立额外的全局变量。
public class MinStack {
/** initialize your data structure here. */
Stack<Integer> stack;
Stack<Integer> minStack;
public MinStack() {
stack = new Stack();
minStack = new Stack();
}
public void push(int x) {
stack.push(x);
if (minStack.isEmpty()) {
minStack.push(x);
} else {
if (x < minStack.peek()) {
minStack.push(x);
} else {
minStack.push(minStack.peek());
}
}
}
public void pop() {
stack.pop();
minStack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
return minStack.peek();
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.getMin();
*/
56, how to use array to implement a queue---BUG FREE
class Queue<T> {
private int front;
private int rear;
private int capacity;
private int size;
T[] queue;
public Queue(int inCapacity) {
capacity = inCapacity;
queue = (T[])new Object[capacity];
front = 0;
rear = 0;
size = 0;
}
public int size() {
return size;
}
public void add(T value) {
if (size == capacity) {
throw new IllegalStateException("Queue is full.");
} else {
queue[rear] = value;
rear = (rear + 1) % capacity;
size++;
}
}
public T remove() {
if (size == 0) {
throw new IllegalStateException("Queue is empty.");
} else {
T res = queue[front];
front = (front + 1) % capacity;
size--;
return res;
}
}
}
55，how to use array to implement a stack---BUG FREE
class Stack<T> {
private int top;
private int capacity;
private T[] stack;
public Stack(int inCapacity) {
capacity = inCapacity;
stack = (T[])new Object[capacity];
top = -1;
}
public void push(T value) {
if (top + 1 == capacity) {
throw new IllegalStateException("Stack is full.");
} else {
stack[++top] = value;
}
}
public T pop() {
if (top == -1) {
throw new IllegalStateException("Stack is empty.");
} else {
return stack[top--];
}
}
public T peek() {
if (top == -1) {
throw new IllegalStateException("Stack is empty.");
} else {
return stack[top];
}
}
}
53,reverse nodes in k-group --- https://leetcode.com/problems/reverse-nodes-in-k-group/---BUG FREE
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. You may not alter the values in the nodes, only nodes itself may be changed. Only constant memory is allowed. For example, Given this linked list: 1->2->3->4->5 For k = 2, you should return: 2->1->4->3->5 For k = 3, you should return: 3->2->1->4->5
CUR: 思路不清晰
答案和思路：找到pre和tail，不断地把pre后面的插入到tail后面。首先找到tail。然后把pre和tail之间的都插入到tail的后面，然后重点是下一次开始是从pre.next，因为他会被插入到最后。所以，需要把它记录下来。
注意break容易出错，一定要在while这里再break一次。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null || head.next == null || k < 1) {
return head;
}
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode pre = dummy;
ListNode tail = dummy;
while (true) {
int count = k;
while (count > 0 && tail != null) {
count--;
tail = tail.next;
}
if (tail == null) {
break;
}
ListNode next = pre.next;
while (pre.next != tail) {
ListNode temp = pre.next;
pre.next = pre.next.next;
temp.next = tail.next;
tail.next = temp;
}
pre = next;
tail = next;
}
return dummy.next;
}
}
52, partition list ---https://leetcode.com/problems/partition-list/---BUG FREE
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. For example, Given 1->4->3->2->5->2 and x = 3, return 1->2->2->4->3->5.
答案：注意不需要建立新节点。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
if (head == null) {
return head;
}
ListNode smallDummy = new ListNode(-1);
ListNode bigDummy = new ListNode(-1);
ListNode small = smallDummy;
ListNode big = bigDummy;
while (head != null) {
if (head.val < x) {
small.next = head;
small = small.next;
} else {
big.next = head;
big = big.next;
}
head = head.next;
}
big.next = null;
small.next = bigDummy.next;
return smallDummy.next;
}
}
51,reorder list---https://leetcode.com/problems/reorder-list/---BUG FREE
Given a singly linked list L: L0→L1→…→Ln-1→Ln, reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→… You must do this in-place without altering the nodes' values.
CUR：在把middle.next给right之前不要先赋值为null。
答案和思路：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public void reorderList(ListNode head) {
if (head == null || head.next == null) {
return;
}
// find the middle
ListNode middle = findMiddle(head);
// reverse the last part
System.out.println(middle.val);
ListNode right = middle.next;
middle.next = null;
ListNode newNode = reverse(right);
// insert
ListNode first = head;
ListNode second = newNode;
while (second != null) {
ListNode temp1 = first.next;
ListNode temp2 = second.next;
first.next = second;
first.next.next = temp1;
second = temp2;
first = temp1;
}
}
public static ListNode findMiddle(ListNode head) {
ListNode fast = head;
ListNode slow = head;
while (fast.next != null && fast.next.next != null) {
fast = fast.next.next;
slow = slow.next;
}
return slow;
}
public static ListNode reverse(ListNode right) {
ListNode newNode = null;
ListNode cur = right;
while (cur != null) {
ListNode temp = cur.next;
cur.next = newNode;
newNode = cur;
cur = temp;
}
return newNode;
}
}
50,insertion sort list---https://leetcode.com/problems/insertion-sort-list/---BUG FREE
答案：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode insertionSortList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode(-1);
ListNode cur = head;
while (cur != null) {
ListNode pre = dummy;
while (pre.next != null && pre.next.val < cur.val) {
pre = pre.next;
}
ListNode temp = pre.next;
pre.next = new ListNode(cur.val);
pre.next.next = temp;
cur = cur.next;
}
return dummy.next;
}
}
49，inserction of two linkedList---https://leetcode.com/problems/intersection-of-two-linked-lists/---BUG FREE
Write a program to find the node at which the intersection of two singly linked lists begins.
For example, the following two linked lists:
A: a1 → a2
↘
c1 → c2 → c3
↗
B: b1 → b2 → b3
begin to intersect at node c1.
答案和分析：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
if (headA == null || headB == null) {
return null;
}
int lenA = getLength(headA);
int lenB = getLength(headB);
if (lenA >= lenB) {
return getNode(headA, headB, lenA, lenB);
} else {
return getNode(headB, headA, lenB, lenA);
}
}
public static ListNode getNode(ListNode max, ListNode min, int maxLen, int minLen) {
ListNode maxNode = max;
ListNode minNode = min;
for (int i = 0; i < maxLen - minLen; i++) {
maxNode = maxNode.next;
}
while (maxNode != null && minNode != null) {
if (maxNode == minNode) {
return maxNode;
}
maxNode = maxNode.next;
minNode = minNode.next;
}
return null;
}
public static int getLength(ListNode head) {
ListNode cur = head;
int len = 0;
while (cur != null) {
len++;
cur = cur.next;
}
return len;
}
}
48,swap nodes in pairs---https://leetcode.com/problems/swap-nodes-in-pairs/--BUG FREE
Given a linked list, swap every two adjacent nodes and return its head. For example, Given ->->->, you should ->->->. Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
CUR：思路不清晰
答案和解析：利用好first，second来辅助。dummy结点
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode pre = dummy;
while (pre.next != null && pre.next.next != null) {
ListNode first = pre.next;
ListNode second = pre.next.next;
ListNode temp = second.next;
pre.next = second;
second.next = first;
first.next = temp;
pre = pre.next.next;
}
return dummy.next;
}
}
47,remove nth node from end of list---https://leetcode.com/problems/remove-nth-node-from-end-of-list/---BUG FREE
Given a linked list, remove the nth node from the end of list and return its head. For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5.
思路：双指针，注意的就是，千万别忘了dummy.next = head;
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if (head == null) {
return null;
}
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode fast = dummy;
ListNode slow = dummy;
for (int i = 0; i < n; i++) {
fast = fast.next;
}
while (fast.next != null) {
fast = fast.next;
slow = slow.next;
}
slow.next = slow.next.next;
return dummy.next;
}
}
46,add two numbers---https://leetcode.com/problems/add-two-numbers/---BUG FREE
CUR:注意value进行取模之前不要先除10.就是算digit和value不要相互影响。统一的办法就是先求和，然后只有存的时候才取余。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(-1);
int value = 0;
int digit = 0;
ListNode cur1 = l1;
ListNode cur2 = l2;
ListNode cur = dummy;
while (cur1 != null && cur2 != null) {
value = cur1.val + cur2.val + digit;
digit = value / 10;
cur.next = new ListNode(value % 10);
cur = cur.next;
cur1 = cur1.next;
cur2 = cur2.next;
}
while (cur1 != null) {
value = cur1.val + digit;
digit = value / 10;
cur.next = new ListNode(value % 10);
cur = cur.next;
cur1 = cur1.next;
}
while (cur2 != null) {
value = cur2.val + digit;
digit = value / 10;
cur.next = new ListNode(value % 10);
cur = cur.next;
cur2 = cur2.next;
}
if (digit != 0) {
cur.next = new ListNode(digit);
}
return dummy.next;
}
}
45，delete node in a linked list---https://leetcode.com/problems/delete-node-in-a-linked-list/---BUG FREE
Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -> 4 after calling your function.
CUR：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public void deleteNode(ListNode node) {
node.val = node.next.val;
node.next = node.next.next;
}
}
44,reverse linked list II---https://leetcode.com/problems/reverse-linked-list-ii/---BUG FREE
CUR:思路混乱
第二天，记住pre.next.next 指向的是pre.next.而不是start.用一个cur来记录当前需要插入的结点，会使程序简洁。
答案和思路：利用dummy辅助，找到start，然后把start后面的不断插入到pre后面。一共进行n-m次操作。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseBetween(ListNode head, int m, int n) {
if (head == null || m <= 0 || n <= 0) {
return head;
}
// find pre
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode pre = dummy;
for (int i = 0; i < m - 1; i++) {
pre = pre.next;
}
ListNode start = pre.next;
ListNode cur = start.next;
for (int i = 0; i < n - m; i++) {
start.next = cur.next;
cur.next = pre.next;
pre.next = cur;
cur = start.next;
}
return dummy.next;
}
}
43, reverse linked list I---https://leetcode.com/problems/reverse-linked-list/--BUG FREE
Reverse a singly linked list.
CUR:思路有些混乱。
答案和思路：其实就是用一个cur指向链表。然后把每一个cur指向的拿下来放到新链表的头。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode cur = head;
ListNode newHead = null;
while (cur != null) {
ListNode temp = cur.next;
cur.next = newHead;
newHead = cur;
cur = temp;
}
return newHead;
}
}
42,merger two sorted list--- https://leetcode.com/problems/merge-two-sorted-lists/---BUG FREE
Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.
答案：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
}
if (l2 == null) {
return l1;
}
ListNode node1 = l1;
ListNode node2 = l2;
ListNode newHead = new ListNode(-1);
ListNode cur = newHead;
while (l1 != null && l2 != null) {
if (l1.val <= l2.val) {
cur.next = new ListNode(l1.val);
l1 = l1.next;
} else {
cur.next = new ListNode(l2.val);
l2 = l2.next;
}
cur = cur.next;
}
if (l1 != null) {
cur.next = l1;
}
if (l2 != null) {
cur.next = l2;
}
return newHead.next;
}
}
41，remove duplicates from sorted list II--- https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/---BUG FREE
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. For example, Given 1->2->3->3->4->4->5, return 1->2->5. Given 1->1->1->2->3, return 2->3.
CUR:思路不清晰。
第二天，一个是思路有卡顿。第二个是：pre.next = pre.next.next出现了误笔。
答案和思路：用一个dummy来辅助。然后pre.next, pre.next.next的用来控制判断。如果重复了，那么就进去循环找只要是等于pre.next.val的都跳过去。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode pre = dummy;
while (pre.next != null && pre.next.next != null) {
if (pre.next.val == pre.next.next.val) {
int val = pre.next.val;
while (pre.next != null && pre.next.val == val) {
pre.next = pre.next.next;
}
} else {
pre = pre.next;
}
}
return dummy.next;
}
}
40,remove duplicates from sorted list I---https://leetcode.com/problems/remove-duplicates-from-sorted-list/--BUG FREE
Given a sorted linked list, delete all duplicates such that each element appear only once. For example, Given 1->1->2, return 1->2. Given 1->1->2->3->3, return 1->2->3.
答案和思路：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode cur = head;
while (cur != null) {
ListNode next = cur.next;
while (next != null && next.val == cur.val) {
cur.next = next.next;
next = next.next;
}
cur = cur.next;
}
return head;
}
}
39, Linked List Cycle II---https://leetcode.com/problems/linked-list-cycle-ii/---BUG FREE
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
答案和思路：
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode detectCycle(ListNode head) {
if (head == null || head.next == null) {
return null;
}
ListNode fast = head;
ListNode slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
if (fast == slow) {
break;
}
}
if (fast != slow) {
return null;
}
fast = head;
while (fast != slow) {
fast = fast.next;
slow = slow.next;
}
return fast;
}
}
38,Linked List Cycle I ---https://leetcode.com/problems/linked-list-cycle/---BUG FREE
Given a linked list, determine if it has a cycle in it.
答案和思路:快慢指针
/**
* Definition for singly-linked list.
* class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public boolean hasCycle(ListNode head) {
if (head == null || head.next == null) {
return false;
}
ListNode fast = head;
ListNode slow = head;
while (fast != null && fast.next != null) {
fast = fast.next.next;
slow = slow.next;
if (fast == slow) {
return true;
}
}
return false;
}
}
37, find the middle node---BUG FREE
class ListNode {
ListNode head;
ListNode next;
int val;
ListNode(int x) {
val = x;
}
}
public class LinkedList {
public static void main(String[] args) {
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
head.next.next.next = new ListNode(4);
head.next.next.next.next = new ListNode(5);
System.out.println(midNode(head).val);
}
public static ListNode midNode(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode slow = head;
ListNode fast = head;
while (fast.next != null && fast.next.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}
36, remove nth node from end of list---https://leetcode.com/problems/remove-nth-node-from-end-of-list/---BUG FREE
Given a linked list, remove the nth node from the end of list and return its head. For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5.
答案和思路：利用双指针先找到他的前一个。然后next.next的方式跳过。因为可能会是删除头结点，所以需要建立dummy结点。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
if (head == null || n <= 0) {
return null;
}
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode first = dummy;
for (int i = 0; i < n; i++) {
if (first == null) {
return head;
}
first = first.next;
}
ListNode second = dummy;
while (first.next != null) {
second = second.next;
first = first.next;
}
second.next = second.next.next;
return dummy.next;
}
}
35，kth node from end--BUG FREE
class ListNode {
ListNode head;
ListNode next;
int val;
ListNode(int x) {
val = x;
}
}
public class LinkedList {
public static void main(String[] args) {
ListNode head = new ListNode(1);
head.next = new ListNode(2);
head.next.next = new ListNode(3);
System.out.println(kthNodeFromEnd(head, 2).val);
}
public static ListNode kthNodeFromEnd(ListNode head, int k) {
ListNode first = head;
ListNode second = head;
for (int i = 0; i < k; i++) {
if (first == null) {
return null;
}
first = first.next;
}
while (first != null) {
first = first.next;
second = second.next;
}
return second;
}
}
34，implement linkedlist---BUG FREE
CUR:思路还不是很清晰。
class ListNode {
ListNode head;
ListNode next;
int val;
ListNode(int x) {
val = x;
}
}
public class LinkedList {
private static ListNode head = null;
private static int length = 0;
public static boolean checkBoundsExclusive(int index) {
if (index < 0 || index >= length) {
System.out.println("ArrayIndexOutOfBoundException");
return true;
} else {
return false;
}
}
public static ListNode getEntry(int index) {
if (!checkBoundsExclusive(index)) {
ListNode cur = head;
for (int i = 0; i < index; i++) {
cur = cur.next;
}
return cur;
} else {
return null;
}
}
public static void set(int index, int value) {
if (!checkBoundsExclusive(index)) {
ListNode node = getEntry(index);
node.val = value;
}
}
public static void add(int index, int value) {
if (!checkBoundsExclusive(index)) {
ListNode newNode = new ListNode(value);
if (index == 0) {
newNode.next = head;
head = newNode;
return;
}
ListNode pre = getEntry(index - 1);
ListNode next = pre.next;
pre.next = newNode;
pre.next.next = next;
}
}
public static void remove(int index) {
if (!checkBoundsExclusive(index)) {
if (index == 0) {
head = head.next;
}
ListNode pre = getEntry(index - 1);
pre.next = pre.next.next;
}
}
}
33, divide two integer-----https://leetcode.com/problems/divide-two-integers/---BUG FREE
Divide two integers without using multiplication, division and mod operator. If it is overflow, return MAX_INT.
CUR：1，思路不清晰。2，边界条件把握不好。3，注意要先变成long才可以用移位来算。并且注意x，y的类型是long。4，最后要判断正负。
public class Solution {
public int divide(int dividend, int divisor) {
if (divisor == 0) {
return dividend >= 0 ? Integer.MAX_VALUE : Integer.MIN_VALUE;
}
if (dividend == 0) {
return 0;
}
if (dividend == Integer.MIN_VALUE && divisor == -1) {
return Integer.MAX_VALUE;
}
boolean isNegative = dividend > 0 && divisor < 0 || dividend < 0 && divisor > 0;
long x = Math.abs((long) dividend);
long y = Math.abs((long) divisor);
int result = 0;
while (x >= y) {
int shift = 0;
while (x >= (y << shift)) {
shift++;
}
result += 1 << (shift - 1);
x = x - (y << (shift - 1));
}
return isNegative ? -result : result;
}
}
32，distinct subsequences-----https://leetcode.com/problems/distinct-subsequences/---BUG FREE
Given a string S and a string T, count the number of distinct subsequences of T in S. A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not). Here is an example: S = "rabbbit", T = "rabbit" Return 3.
CUR:思路不清晰。
CUR:思路不清晰。而且string.length是错的。
思路：dp[i][j]表示的是s的前i个字符里面t的前j个字符的序列在里面有多少个。那么，如果s的第i个和t的第j个字符相等的话，就可以==dp[i - 1][j - 1] + dp[i - 1][j]。否则，就只==dp[i - 1][j]。
public class Solution {
public int numDistinct(String s, String t) {
if (s == null) {
return 0;
}
if (t == null) {
return 1;
}
int lenS = s.length();
int lenT = t.length();
int[][] dp = new int[lenS + 1][lenT + 1];
// 前lenS，前lenT里面有多少个。
for (int i = 0; i < lenS; i++) {
dp[i][0] = 1;
}
for (int i = 1; i <= lenS; i++) {
for (int j = 1; j <= lenT; j++) {
if (s.charAt(i - 1) == t.charAt(j - 1)) {
dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
} else {
dp[i][j] = dp[i - 1][j];
}
}
}
return dp[lenS][lenT];
}
}
31,best time to buy and sell stock-----https://leetcode.com/problems/best-time-to-buy-and-sell-stock/---bug free
Say you have an array for which the ith element is the price of a given stock on day i. If you were only permitted to complete at most one transaction (ie, buy one and sell one share of the stock), design an algorithm to find the maximum profit. Example 1: Input: [7, 1, 5, 3, 6, 4] Output: 5 max. difference = 6-1 = 5 (not 7-1 = 6, as selling price needs to be larger than buying price) Example 2: Input: [7, 6, 4, 3, 1] Output: 0 In this case, no transaction is done, i.e. max profit = 0.
答案和思路：
public class Solution {
public int maxProfit(int[] prices) {
if (prices == null || prices.length == 0) {
return 0;
}
int minPrice = prices[0];
int maxProfit = 0;
for (int i = 1; i < prices.length; i++) {
minPrice = Math.min(minPrice, prices[i]);
maxProfit = Math.max(maxProfit, prices[i] - minPrice);
}
return maxProfit;
}
}
30,Triangle----https://leetcode.com/problems/triangle/---BUG FREE
iven a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below.
For example, given the following triangle
[
[2],
[3,4],
[6,5,7],
[4,1,8,3]
]
The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11).
CUR:直接在list上面进行set。先把他get出来再set。
第二天：i++误写成i--。然后j<= i这里也记不清。
public class Solution {
public int minimumTotal(List<List<Integer>> triangle) {
if (triangle == null || triangle.size() == 0) {
return 0;
}
int size = triangle.size();
for (int i = size - 2; i >= 0; i--) {
// find the min
for (int j = 0; j <= i; j++) {
triangle.get(i).set(j, triangle.get(i).get(j) + Math.min(triangle.get(i + 1).get(j), triangle.get(i + 1).get(j + 1)));
}
}
return triangle.get(0).get(0);
}
}
29，Maximum subarray------https://leetcode.com/problems/maximum-subarray/---bug free
Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [−2,1,−3,4,−1,2,1,−5,4], the contiguous subarray [4,−1,2,1] has the largest sum = 6.
答案：
Find the contiguous subarray within an array (containing at least one number) which has the largest sum. For example, given the array [−2,1,−3,4,−1,2,1,−5,4], the contiguous subarray [4,−1,2,1] has the largest sum = 6.
滚动指针优化：
public class Solution {
public int maxSubArray(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
int[] dp = new int[2];
int max = nums[0];
dp[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
dp[i % 2] = Math.max(dp[(i - 1) % 2] + nums[i], nums[i]);
max = Math.max(max, dp[i % 2]);
}
return max;
}
}
28, Climbing Stairs-----https://leetcode.com/problems/climbing-stairs/---bug free
You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Subscribe to see which companies asked this question
答案：
public class Solution {
public int climbStairs(int n) {
if (n <= 0) {
return 0;
}
if (n <= 2) {
return n;
}
int[] dp = new int[n];
dp[0] = 1;
dp[1] = 2;
for (int i = 2; i < n; i++) {
dp[i] = dp[i - 1] + dp[i - 2];
}
return dp[n - 1];
}
}
滚动指针优化：
public class Solution {
public int climbStairs(int n) {
if (n <= 0) {
return 0;
}
if (n <= 2) {
return n;
}
int[] dp = new int[2];
dp[0] = 1;
dp[1] = 2;
for (int i = 2; i < n; i++) {
dp[i % 2] = dp[(i - 1) % 2] + dp[(i - 2) % 2];
}
return dp[(n - 1) % 2];
}
}
27，Minimum Path Sum-----https://leetcode.com/problems/minimum-path-sum/---BUG FREE
Given a m x n grid filled with non-negative numbers, find a path from top left to bottom right which minimizes the sum of all numbers along its path. Note: You can only move either down or right at any point in time.
CUR:总是忘记了要加上grid[i][j]。无论是初始化还是计算的时候都要加上去。
答案和思路：
public class Solution {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[m][n];
dp[0][0] = grid[0][0];
for (int i = 1; i < m; i++) {
dp[i][0] = dp[i - 1][0] + grid[i][0];
}
for (int j = 1; j < n; j++) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + grid[i][j];
}
}
return dp[m - 1][n - 1];
}
}
滚动数组来优化空间复杂度：
public class Solution {
public int minPathSum(int[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int[][] dp = new int[2][n];
dp[0][0] = grid[0][0];
// for (int i = 1; i < m; i++) {
// dp[i][0] = dp[i - 1][0] + grid[i][0];
// }
for (int j = 1; j < n; j++) {
dp[0][j] = dp[0][j - 1] + grid[0][j];
}
for (int i = 1; i < m; i++) {
dp[i % 2][0] = dp[(i - 1) % 2][0] + grid[i][0];
for (int j = 1; j < n; j++) {
dp[i % 2][j] = Math.min(dp[(i - 1) % 2][j], dp[i % 2][j - 1]) + grid[i][j];
}
}
return dp[(m - 1) % 2][n - 1];
}
}
26(2),Unique Path II-----https://leetcode.com/problems/unique-paths-ii/---BUG FREE
Follow up for "Unique Paths": Now consider if some obstacles are added to the grids. How many unique paths would there be? An obstacle and empty space is marked as 1 and 0 respectively in the grid. For example, There is one obstacle in the middle of a 3x3 grid as illustrated below. [ [0,0,0], [0,1,0], [0,0,0] ] The total number of unique paths is 2.
题目大意：多了障碍物。
解法一：
CUR：注意，初始化的时候，一旦发现有一个障碍物，后面的就要都为0了。
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid.length == 0) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] dp = new int[m][n];
// initialize
for (int i = 0; i < m; i++) {
if (obstacleGrid[i][0] != 1) {
dp[i][0] = 1;
} else {
break;
}
}
for (int j = 0; j < n; j++) {
if (obstacleGrid[0][j] != 1) {
dp[0][j] = 1;
} else {
break;
}
}
// calculate
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] != 1) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
}
return dp[m - 1][n - 1];
}
}
解法二：滚动数组优化dp。注意，因为是滚动数组了。所以每一次if之后都要用else清0.
CUR：
public class Solution {
public int uniquePathsWithObstacles(int[][] obstacleGrid) {
if (obstacleGrid == null || obstacleGrid.length == 0) {
return 0;
}
int m = obstacleGrid.length;
int n = obstacleGrid[0].length;
int[][] dp = new int[2][n];
for (int j = 0; j < n; j++) {
if (obstacleGrid[0][j] != 1) {
dp[0][j] = 1;
} else {
break;
}
}
// calculate
for (int i = 1; i < m; i++) {
if (dp[(i - 1) % 2][0] != 0 && obstacleGrid[i][0] != 1) {
dp[i % 2][0] = 1;
} else {
dp[i % 2][0] = 0;
}
for (int j = 1; j < n; j++) {
if (obstacleGrid[i][j] != 1) {
dp[i % 2][j] = dp[(i - 1) % 2][j] + dp[i % 2][j - 1];
} else {
dp[i % 2][j] = 0;
}
}
}
return dp[(m - 1) % 2][n - 1];
}
}
26(1),Unique Paths I-----https://leetcode.com/problems/unique-paths/---BUG FREE
A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below).
答案：
public class Solution {
public int uniquePaths(int m, int n)
{
int[][] dp = new int[m][n];
for (int i = 0; i < n; i++) {
dp[0][i] = 1;
}
for (int i = 0; i < m; i++) {
dp[i][0] = 1;
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
}
return dp[m - 1][n - 1];
}
}
滚动指针来优化空间复杂度：
public class Solution {
public int uniquePaths(int m, int n)
{
int[][] dp = new int[2][n];
for (int i = 0; i < n; i++) {
dp[0][i] = 1;
}
dp[0][0] = 1;
for (int i = 1; i < m; i++) {
dp[i % 2][0] = 1;
for (int j = 1; j < n; j++) {
dp[i % 2][j] = dp[(i - 1) % 2][j] + dp[i % 2][j - 1];
}
}
return dp[(m - 1) % 2][n - 1];
}
}
23, search a 2d matrix II -----https://leetcode.com/problems/search-a-2d-matrix-ii/---BUG FREE
CUR: 思路不清晰。
答案和思路：从左下角开始走，如果>target,就往上走，如果小于就往右走。
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0) {
return false;
}
int m = matrix.length;
int n = matrix[0].length;
int x = m - 1;
int y = 0;
// 从左下角开始出发。
while (x >= 0 && y < n) {
if (matrix[x][y] == target) {
return true;
} else if (matrix[x][y] < target) {
y++;
} else if (matrix[x][y] > target) {
x--;
}
}
return false;
}
}
22,search a 2d matrix I-----https://leetcode.com/problems/search-a-2d-matrix/----BUG FREE
CUR:matrix 误写成nums，down误写成right。row误写成result。所以，一定要改掉这种代码手诟病。
答案和思路：
public class Solution {
public boolean searchMatrix(int[][] matrix, int target) {
if (null == matrix || matrix.length == 0) {
return false;
}
// find the row
int up = 0;
int down = matrix.length - 1;
int row = 0;
while (up <= down) {
int mid = up + (down - up) / 2;
if (matrix[mid][0] == target) {
return true;
} else if (matrix[mid][0] < target) {
row = mid;
up = mid + 1;
} else {
down = mid - 1;
}
}
int left = 0;
int right = matrix[0].length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (matrix[row][mid] == target) {
return true;
} else if (matrix[row][mid] > target) {
right = mid - 1;
} else {
left = mid + 1;
}
}
return false;
}
}
21(2)find minimum in rotated sorted array II-----https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/----BUG FREE
总结：在循环数组里面找minimum，统一跟right比较决定往哪边走。如果没有重复元素,nums[mid] > nums[right], 往右走left = mid + 1, 否则，往左走right = mid - 1。有重复元素的时候，nums[mid] > nums[right],往右走，nums[mid] < nums[right],往左走，right = mid - 1，多了一种情况right--；
循环数组题的总结：判断往哪边走，只需要把mid与right比较就可以了。然后有重复元素的时候，多判断一种情况。
CUR:要用更好的思路来解。虽然因为有重复元素已经退化到了最坏情况是O（n）。但是二分还是可以接着用的。
第二天：思路还是不是很清晰。
1,O（N）：
public class Solution {
public int findMin(int[] nums) {
if (null == nums || nums.length == 0) {
return -1;
}
int min = nums[0];
for (int i = 0; i < nums.length; i++) {
min = Math.min(min, nums[i]);
}
return min;
}
}
二分的思路：1，保证left一直都是存可能结果。但是这里该往那边走，跟mid比较的是right。
2，binary research
public class Solution {
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
if (nums.length == 1 || nums[0] < nums[nums.length - 1]) {
return nums[0];
}
int left = 0;
int right = nums.length - 1;
while (left <= right) {
if (nums[left] < nums[right]) {
return nums[left];
}
int mid = left + (right - left) / 2;
if (mid > 0 && nums[mid] < nums[mid - 1]) {
return nums[mid];
}
if (nums[mid] > nums[right]) {
left = mid + 1;
} else if (nums[mid] < nums[right]) {
right = mid - 1;
} else {
right--;
}
}
return nums[right + 1];
}
}
21(1),find minimum in rotated sorted array I---https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/------BUG FREE
Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). Find the minimum element. You may assume no duplicate exists in the array. Subscribe to see which companies asked this question
CUR:思路不清晰。
第二天：1，忘记进来的时候判断一下是否是严格递增了。
答案和思路：1，首先，要注意的是要特判如果第一个就比最后一个小那么就直接找到了。2，其次是要注意分界的走向还是要根据mid是否位于哪一段上坡线上面。3,因为left,right是不断地变化的，所以，还是要每次进来判断一下是否nums[left] <= nums[right]。如果是的话，直接return。总结，核心思想也是通过来判断mid在哪一个上升坡上来判断往哪边走。如果nums[mid] >= nums[left], 往右走，否则往左走。
public class Solution {
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
if (nums[0] <= nums[nums.length - 1]) {
return nums[0];
}
int left = 0;
int right = nums.length - 1;
while (left <= right) {
if (nums[left] <= nums[right]) {
return nums[left];
}
int mid = left + (right - left) / 2;
if (mid > 0 && nums[mid] < nums[mid - 1]) {
return nums[mid];
}
if (nums[right] < nums[mid]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
return -1;
}
}
20,Sqrt(x)-----https://leetcode.com/problems/sqrtx/---BUG FREE
. Sqrt(x) QuestionEditorial Solution My Submissions Total Accepted: Total Submissions: Difficulty: Medium Implement int sqrt(int x). Compute and return the square root of x.
CUR:判断错误。x/mid < mid 时保存result的做法是错的。
第二天：x == 0没有特判。注意，x <= 1return x就可以了。
第三天：left=0；是错的，会导致除数为0，left要从1开始。
答案和思路：注意特判0,1因为right取的是x/2. x / mid == mid, return mid. x/mid > mid, keep the mid as the result.
public class Solution {
public int mySqrt(int x) {
if (x <= 1) {
return x;
}
int left = 1;
int right = x / 2;
int result = 0;
while (left <= right) {
int mid = left + (right - left) / 2;
if (x / mid == mid) {
return mid;
}
if (x / mid > mid) {
result = mid;
left = mid + 1;
} else {
right = mid - 1;
}
}
return result;
}
}
18(2)，Search in rotated sorted Array II-------- https://leetcode.com/problems/search-in-rotated-sorted-array-ii/-----NOT BUG FREE
总结：只需要跟right比就可以。但凡是遇到循环数组查找元素的题，只需要把它找到它位于哪一条严格上升的线上。没有重复元素的时候，只需要比较mid与left之间的大小就可以：<=, > 两种情况。如果有重复元素的话：需要比较：mid > left,左上升曲线， mid < right, 右上升曲线，除此之外，只能判断一下left，然后left++；
Follow up for "Search in Rotated Sorted Array": What if duplicates are allowed? Would this affect the run-time complexity? How and why? Write a function to determine if a given target is in the array.
CUR:思路不清晰。
第二天：1， 把nums[right]误写成right；2，else那里之所以写判断，是因为要left++了。如果left就是要找的值，那么就错过了。
第三天：target < nums[right],应该是<=
思路和答案：1，这里的关键点就是因为有重复元素，导致了他可能处于既比左边大又比右边大，那么就不好找严格的单调区间，所以，需要分三种情况，一个是大于了左边，一个是小于了右边，else。注意，最后一种情况要特判是否等于left，因为不判断，可能退化到O(n).。
public class Solution {
public boolean search(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return false;
}
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return true;
}
if (nums[mid] > nums[right]) {
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else if (nums[mid] < nums[right]) {
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
} else {
if (nums[right] == target) {
return true;
}
right--;
}
}
return false;
}
}
18（1），Search in Rotated Sorted Array-----https://leetcode.com/problems/search-in-rotated-sorted-array/----- BUG FREE
Suppose a sorted array is rotated at some pivot unknown to you beforehand. (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). You are given a target value to search. If found in the array return its index, otherwise return -1. You may assume no duplicate exists in the array.
答案和分析：关键点就是首先看mid位置与left进行比较。循环数组题的总结：判断往哪边走，只需要把mid与right比较就可以了。然后有重复元素的时候，多判断一种情况。
public class Solution {
public int search(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int left = 0;
int right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return mid;
}
if (nums[mid] < nums[right]) {
if (nums[mid] < target && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
} else {
if (nums[left] <= target && target < nums[mid]) {
right = mid - 1;
} else {
left = mid + 1;
}
}
}
return -1;
}
}
17,find insertion -----BUG FREE
Insertion Position Instead of giving -1 when search fails, give the index that the number could be inserted and maintain the order Copyright ©BUG-FREE-For Dream的更多相关文章
- 马化腾:办公用QQ休闲用微信[Dream Catchers论坛]
近日,香港大学举办以创新创业为主题的Dream Catchers论坛.其中腾讯董事局主席马化腾在下午两点四十五分在李兆基会议中心做了专题演讲,分享了自己的创业经历并回答了媒体人张力奋有关产品.整整对手 ...
- 5 Common Interview Mistakes that Could Cost You Your Dream Job (and How to Avoid Them)--ref
There have been many articles on our site on software testing interviews. That is because, we, as IT ...
- <Dare To Dream>团队项目用户验收评审
实验十二 团队作业8—团队项目用户验收评审 任务1:团队作业Beta冲刺 Beta冲刺第一天:http://www.cnblogs.com/Dare-To-Dream/p/9226994.html B ...
- How to Land your Dream Job
今天在code school上面看到一个外国人写的文章,写的很棒,瞧一下外国人的思维和我们有什么不同?分享给所有的朋友 You know how some things ar ...
- POJ 2411 Mondriaan's Dream 插头dp
题目链接: http://poj.org/problem?id=2411 Mondriaan's Dream Time Limit: 3000MSMemory Limit: 65536K 问题描述 S ...
- 改进意见的答复及bug重现
各组对本组的互评链接如下 Thunder:http://www.cnblogs.com/vector121/p/7905300.html 王者荣耀交流协会:http://www.cnblogs.com ...
- Tomcat一个BUG造成CLOSE_WAIT
之前应该提过,我们线上架构整体重新架设了,应用层面使用的是Spring Boot,前段日子因为一些第三方的原因,略有些匆忙的提前开始线上的内测了.然后运维发现了个问题,服务器的HTTPS端口有大量的C ...
- a标签点击跳转失效--IE6、7的奇葩bug
一般运用a标签包含img去实现点击图片跳转的功能,这是前端经常要用到的东西. 今天遇到个神奇的bug:如果在img上再包裹一层div,而且div设置了width和height,则图片区域点击时,无任何 ...
- 关于 Chrome 浏览器中 onresize 事件的 Bug
我在写插件时用到了 onresize 事件,在反复地测试后发现该事件在 Chrome 及 Opera(内核基本与 Chrome 相同,以下统称 Chrome)浏览器打开时就会执行,这种情况也许不能算作 ...
- Chrome出了个小bug：论如何在Chrome下劫持原生只读对象
Chrome出了个小bug:论如何在Chrome下劫持原生只读对象 概述 众所周知,虽然JavaScript是个很灵活的语言,浏览器里很多原生的方法都可以随意覆盖或者重写,比如alert.但是为了保证 ...
随机推荐
- [LeetCode] Binary Tree Level Order Traversal II 二叉树层序遍历之二
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left ...
- javascrip for 循环数组事件绑定
//radio选择填充,字段 var radioArray = ["F_RegisterCode", "F_Subjection", "F_Mecha ...
- POJ2096 Collecting Bugs
Time Limit: 10000MS Memory Limit: 64000K Total Submissions: 5090 Accepted: 2529 Case Time Limit: ...
- 内存VSS/RSS/PSS/USS名词解释
VSS(virtual set size)虚拟耗用内存(包含共享库占用的内存) RSS(Resident set size)实际使用物理内存(包含共享库占用的内存) RSS是进程实际驻存在物理内存的部 ...
- RocketMQ原理解析－Producer
producer producer 1.启动流程 Producer如何感知要发送消息的broker即brokerAddrTable中的值是怎么获得的, 1. 发送消息的时候指定会指定topic,如果 ...
- app xml报错
如下图所示红色区域,xml中第7行提示以下错误,检查了好久,没发现什么异常,请高人指点下,谢谢! error parsing xml:not well-formed (invalid token)
- 将list集合的元素按照添加顺序的倒序进行排列取出
1.方法 Collections.reverse(list); 2.代码示例 /** * 从redis中将现场状态的记录全部取出 * @param aucId * @return */ @Reques ...
- XPath 学习二： 语法
XPath 使用路径表达式来选取 XML 文档中的节点或节点集.节点是通过沿着路径 (path) 或者步 (steps) 来选取的. 下面列出了最有用的路径表达式: 表达式 描述 nodename 选 ...
- IScroll5兼容IE修改
水平不到家,无法像js大神那样讲得头头是道.仅做记录,以备后查. iScroll5是不兼容IE低版本的.为兼容IE低版本(以IE8为例),需做以下工作: 1.事件绑定方式兼容 众所周知,独特的IE有它 ...
- H5+CSS3知识点
概要:CSS3美化样式.自定义字体图标.滤镜设置.CSS3选择器.transform2D转换.新增表单控件.vaild表单验证.表单样式美化等. 属性选择器: E[attr]只使用属性名,但没有确定任 ...