九章lintcode作业题
1 - 从strStr谈面试技巧与代码风格
13.字符串查找
要求：如题
思路：（自写AC）双重循环，内循环读完则成功
还可以用Rabin，KMP算法等
public int strStr(String source, String target) {
if (source == null || target == null) {
return -1;
}
char[] sources = source.toCharArray();
char[] targets = target.toCharArray();
int m = sources.length;
int n = targets.length;
for (int i = 0; i <= m - n; i++) {
int j = 0;
while (j < n && sources[i + j] == targets[j]) {
j++;
}
if (j == n) {
return i;
}
}
return -1;
}
2017-01-16
594.StrStr II
要求：时间复杂度为O(m+n)
思路：用Rabin Karp算法，这种算法沿用了暴力法，但加入了hashcode的思想，因为target和substring的hashcode进行对比只需要O(1)时间复杂度，而且substring变成下一个子串也只需要进行一次O(1)的运算（就是把头部减掉，再加上新的字符的hashcode），等target和substring的hashcode相等再对它们进行真正的字符串比较
warn：由于hashcode大小有可能越界，所以要对其进行一个固定BASE的取模运算，如果变成负的时候要把它加一个BASE变回正数
public class Solution {
public final int BASE = 1000000;
/**
* @param source a source string
* @param target a target string
* @return an integer as index
*/
public int strStr2(String source, String target) {
if (source == null || target == null) {
return -1;
}
int m = target.length();
if (m == 0) {
return 0;
}
// count 31 ^ m
int power = 1;
for (int i = 0; i < m; i++) {
power = (power * 31) % BASE;
}
// target hashcode
int targetCode = 0;
for (int i = 0; i < m; i++) {
targetCode = (targetCode * 31 + target.charAt(i)) % BASE;
}
// substring hashcode
int substrCode = 0;
for (int i = 0; i < source.length(); i++) {
substrCode = (substrCode * 31 + source.charAt(i)) % BASE;
if (i < m - 1) {
continue;
}
// 超过m长度后减去头字符的hashcode
if (i >= m) {
substrCode = substrCode - (power * source.charAt(i - m)) % BASE;
//当前字符的hashcode比之前取模后的字符串的hashcode大就变成负数
if (substrCode < 0) {
substrCode += BASE;
}
}
if (substrCode == targetCode) {
if (source.substring(i - m + 1, i + 1).equals(target)) {
return i - m + 1;
}
}
}
return -1;
}
}
2017-09-19
17.子集
要求：集合中不含重复，答案中集合的元素升序排列
思路：（自写修改AC）DFS模板，背
warn：记得deep copy
public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
ArrayList<ArrayList<Integer>> results = new ArrayList<>();
if (nums == null || nums.length == 0) {
return results;
}
ArrayList<Integer> subset = new ArrayList<>();
//排序非必要
Arrays.sort(nums);
//从空集开始
dfsHelper(nums, 0, subset, results);
return results;
}
private void dfsHelper(int[] nums,
int startIndex,
ArrayList<Integer> subset,
ArrayList<ArrayList<Integer>> results) {
//deep copy
results.add(new ArrayList<Integer>(subset));
for (int i = startIndex; i < nums.length; i++) {
subset.add(nums[i]);
dfsHelper(nums, i + 1, subset, results);
subset.remove(subset.size() - 1);
}
}
2017-01-16
18.带重复元素的子集
要求：集合中有重复元素，不能出现重复的子集
思路：模板中加入了nums[i] == nums[i - 1] && i != startIndex，表示新一轮从startIndex开始加入的不能重复
public class Solution {
/*
* @param nums: A set of numbers.
* @return: A list of lists. All valid subsets.
*/
public List<List<Integer>> subsetsWithDup(int[] nums) {
List<List<Integer>> results = new ArrayList<>();
if (nums == null || nums.length == 0) {
return results;
}
Arrays.sort(nums);
List<Integer> subset = new ArrayList<>();
dfsHelper(nums, 0, subset, results);
return results;
}
private void dfsHelper(int[] nums,
int startIndex,
List<Integer> subset,
List<List<Integer>> results) {
results.add(new ArrayList<>(subset));
for (int i = startIndex; i < nums.length; i++) {
if (i != 0 && nums[i] == nums[i - 1] && i != startIndex) {
continue;
}
subset.add(nums[i]);
dfsHelper(nums, i + 1, subset, results);
subset.remove(subset.size() - 1);
}
}
}
2017-09-19
15.全排序
要求：没有重复元素
思路：与子集一样，但要加入判断数组visited[]，而且还要判断每次生成的list长度是否等于列表长度才加入结果中
2017-09-19
16.带有重复元素的排序
要求：如题
思路：与含有重复元素的子集一样，要判断当前的数是否与上一个数一样，并且判断上一个数是否已放在排序中
class Solution {
/**
* @param nums: A list of integers.
* @return: A list of unique permutations.
*/
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> lists = new ArrayList<>();
List<Integer> list = new ArrayList<>();
if (nums == null) {
return lists;
}
if (nums.length == 0) {
lists.add(list);
return lists;
}
Arrays.sort(nums);
int[] visited = new int[nums.length];
helper(nums, visited, list, lists);
return lists;
}
private void helper(int[] nums, int[] visited,
List<Integer> list, List<List<Integer>> lists) {
if (list.size() == nums.length) {
lists.add(new ArrayList<Integer>(list));
}
for (int i = 0; i < nums.length; i++) {
if (visited[i] == 1) {
continue;
}
if (i != 0 && nums[i] == nums[i - 1] && visited[i - 1] == 0) {
continue;
}
visited[i] = 1;
list.add(nums[i]);
helper(nums, visited, list, lists);
list.remove(list.size() - 1);
visited[i] = 0;
}
}
}
2017-09-19
2 - 二分法与排序数组
457.二分查找标准写法
注意：循环判断条件、double check
public int findPosition(int[] nums, int target) {
// write your code here
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
return mid;
} else if (nums[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (nums[start] == target) {
return start;
}
if (nums[end] == target) {
return end;
}
return -1;
}
2018-01-17
458.目标最后位置
要求：二分法找目标最后一次
思路：（自写但有错误）double check时候要把end放在前面
public int lastPosition(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
start = mid;
} else if (nums[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (nums[end] == target) {
return end;
}
if (nums[start] == target) {
return start;
}
return -1;
}
2017-01-23
141.x的平方根
要求：二分法找
思路：（自写有错误）定义时应为long，防止溢出
public int sqrt(int x) {
if (x == 0) {
return 0;
}
long start = 1, end = x;
while (start + 1 < end) {
long mid = start + (end - start) / 2;
if (mid * mid == x) {
return (int) mid;
} else if (mid * mid < x) {
start = mid;
} else {
end = mid;
}
}
return (int) start;
}
2017-01-23
14.二分查找
要求：找第一次出现的位置
思路：（自写一次AC）注意dc时候把start放前面
public int binarySearch(int[] nums, int target) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] == target) {
end = mid;
} else if (nums[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (nums[start] == target) {
return start;
}
if (nums[end] == target) {
return end;
}
return -1;
}
2017-01-23
447.在大数组中查找
要求：只能通过get来获得一个位置，不知道数组长度，获取target第一次出现位置
思路：（自写一次AC）多加了一个找长度的while
public int searchBigSortedArray(ArrayReader reader, int target) {
//找length
int length = 1;
while (reader.get(length - 1) < target) {
length *= 2;
}
int start = 0, end = length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (reader.get(mid) == target) {
end = mid;
} else if (reader.get(mid) < target) {
start = mid;
} else {
end = mid;
}
}
if (reader.get(start) == target) {
return start;
}
if (reader.get(end) == target) {
return end;
}
return -1;
}
2017-01-23
183.木材加工
要求：找满足数量k的最长分法
思路：（自写出错）二分答案，mid选的是长度，判断的是分了之后的结果是否满足k，自写时没加入count方法所以double check出错
public int woodCut(int[] L, int k) {
if (L == null || L.length == 0) {
return 0;
}
Arrays.sort(L); //也可以遍历求最大值
int start = 1, end = L[L.length - 1];
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (count(L, mid) >= k) {
start = mid;
} else {
end = mid;
}
}
if (count(L, start) >= k) {
return start;
}
if (count(L, end) >= k) {
return end;
}
return 0;
}
//count方法求每次的数量
private int count(int[] L, int len) {
int sum = 0;
for (int i = 0; i < L.length; i++) {
sum += L[i] / len;
}
return sum;
}
2017-01-23
159.寻找旋转排序数组(RSA)中的最小值
要求：升序数组旋转后，求旋转的那个最小值点
思路：（掌握思路后自写一次AC）选最后一个数作为target，然后求小于等于target的第一个位置上的数
public int findMin(int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
int start = 0, end = nums.length - 1;
//choose last number as target
//and find the first position if nums[i] <= target
int target = nums[nums.length - 1];
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] <= target) {
end = mid;
} else {
start = mid;
}
}
//double check
if (nums[start] <= target) {
return nums[start];
} else {
return nums[end];
}
}
2017-01-24
75.寻找峰值
要求：找出任意峰值点
思路：（掌握思路一次AC）与旁边的比较，if先写start = mid，else为end = mid，最后要dc，属于Half Half法
tips：end取length - 2
public int findPeak(int[] A) {
if (A == null || A.length == 0) {
return -1;
}
int start = 0, end = A.length - 2; //peak can only happen before len - 2
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (A[mid] < A[mid + 1]) {
start = mid;
} else {
end = mid;
}
}
if (A[start] < A[end]) {
return end;
} else {
return start;
}
}
2017-01-24
74.第一个错误的代码版本
要求：出现错误后就一直错，用接口判断是否错
思路：（读清楚题后AC）XXOO找第一个
public int findFirstBadVersion(int n) {
if (n == 0) {
return 0;
}
int start = 1, end = n;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (SVNRepo.isBadVersion(mid)) {
end = mid;
} else {
start = mid;
}
}
if (SVNRepo.isBadVersion(start)) {
return start;
} else {
return end;
}
}
2017-01-25
62.搜索旋转排序数组
要求：在RSA中找出target
思路：（答案）属于Half Half，用start来判断mid在断点左还是右，左的时候利用start和mid夹逼，右的时候用mid和end夹
public int search(int[] A, int target) {
if (A == null || A.length == 0) {
return -1;
}
int start = 0, end = A.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (A[mid] == target) {
return mid;
}
if (A[start] < A[mid]) {
if (A[start] <= target && target <= A[mid]) {
end = mid;
} else {
start = mid;
}
} else {
if (A[mid] <= target && target <= A[end]) {
start = mid;
} else {
end = mid;
}
}
}
if (A[start] == target) {
return start;
}
if (A[end] == target) {
return end;
}
return -1;
}
2017-01-25
437.书籍复印copy books
要求：pages数组表示每本书的页数，求k个人复印的最短时间
思路：（答案，还不完全理解）二分答案，先算出pages最大和总数，最快的情况是最大的复印完就全部复印完，最坏的情况是只有一个人复印全部书，所以是总数，没看懂countCopiers方法
public int copyBooks(int[] pages, int k) {
if (pages.length == 0) {
return 0;
}
int total = 0;
int max = pages[0];
for (int i = 0; i < pages.length; i++) {
total += pages[i];
if (max < pages[i]) {
max = pages[i];
}
}
int start = max;
int end = total;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (countCopiers(pages, mid) > k) {
start = mid;
} else {
end = mid;
}
}
if (countCopiers(pages, start) <= k) {
return start;
} else {
return end;
}
}
private int countCopiers(int[] pages, int limit) {
int copiers = 1;
int sum = pages[0];
for (int i = 1; i < pages.length; i++) {
if (sum + pages[i] > limit) {
copiers++;
sum = 0;
}
sum += pages[i];
}
return copiers;
}
2017-01-25
459.closest number in sorted array
要求：找出排序数组中与target最接近的数，不要求第一个还是最后一个
思路：（自写AC）在模板基础上加多一个判断，记得double check
public int closestNumber(int[] A, int target) {
if (A == null || A.length == 0) {
return -1;
}
int start = 0, end = A.length - 1;
int diff = Math.abs(A[0] - target);
int index = 0;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (Math.abs(A[mid] - target) < diff) {
diff = Math.abs(A[mid] - target);
index = mid;
}
if (A[mid] == target) {
return mid;
} else if (A[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (Math.abs(A[start] - target) < diff) {
return start;
}
if (Math.abs(A[end] - target) < diff) {
return end;
}
return index;
}
2017-03-08
28.search a 2D matrix
要求：在二维矩阵查找一个数是否存在
思路：（自写AC）找右上或左下
public boolean searchMatrix(int[][] matrix, int target) {
if (matrix == null || matrix.length == 0 || matrix[0].length == 0) {
return false;
}
int row = 0, col = matrix[0].length - 1;
while (row < matrix.length && col >= 0) {
if (matrix[row][col] == target) {
return true;
} else if (matrix[row][col] < target) {
row++;
} else {
col--;
}
}
return false;
}
2017-03-08
38.search a 2D matrix II
要求：找出target出现几次
思路：和上题一样，只不过相等时要行与列同时变化
2017-09-25
585.maximum number in mountain sequence
要求：序列先升后降 找顶点，与75题比较差别是这个序列中只存在一个峰值，而75可能有多个
思路：（自写AC）与旁边比较
public int mountainSequence(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
int start = 0, end = nums.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (nums[mid] > nums[mid - 1] && nums[mid] > nums[mid + 1]) {
return nums[mid];
} else if (nums[mid] > nums[mid - 1]) {
start = mid;
} else {
end = mid;
}
}
if (nums[start] > nums[start + 1]) {
return nums[start];
} else {
return nums[end];
}
}
2017-03-08
600.smallest rectangle enclosing black pixels(hard)
要求：image矩阵中包括1、0，其中1互相连同且image中只有一块，给出其中一个1的坐标，求把1的区域围绕起来的最小矩形
思路：（答案，没自写，答案的style不好）二分法，把给出坐标的四个方向都做一次二分，求出上下左右
public int minArea(char[][] image, int x, int y) {
// Write your code here
int m = image.length;
if (m == 0)
return 0;
int n = image[0].length;
if (n == 0)
return 0;
int start = y;
int end = n - 1;
int mid;
while (start < end) {
mid = start + (end - start) / 2 + 1;
if (checkColumn(image, mid)) {
start = mid;
} else {
end = mid - 1;
}
}
int right = start;
start = 0;
end = y;
while (start < end) {
mid = start + (end - start) / 2;
if (checkColumn(image, mid)) {
end = mid;
} else {
start = mid + 1;
}
}
int left = start;
start = x;
end = m - 1;
while (start < end) {
mid = start + (end - start) / 2 + 1;
if (checkRow(image, mid)) {
start = mid;
} else {
end = mid - 1;
}
}
int down = start;
start = 0;
end = x;
while (start < end) {
mid = start + (end - start) / 2;
if (checkRow(image, mid)) {
end = mid;
} else {
start = mid + 1;
}
}
int up = start;
return (right - left + 1) * (down - up + 1);
}
private boolean checkColumn(char[][] image, int col) {
for (int i = 0; i < image.length; i++) {
if (image[i][col] == '1') {
return true;
}
}
return false;
}
private boolean checkRow(char[][] image, int row) {
for (int j = 0; j < image[0].length; j++) {
if (image[row][j] == '1') {
return true;
}
}
return false;
}
还可以用bfs来做！
2017-03-08
462.目标出现总和
要求：找出target出现的次数，数组为升序
思路（自写）：二分，找到target后用left与right指针找出第一次发现后，两边分别还有多少个相同的数
public class Solution {
/*
* @param A: A an integer array sorted in ascending order
* @param target: An integer
* @return: An integer
*/
public int totalOccurrence(int[] A, int target) {
// write your code here
if (A == null || A.length == 0) {
return 0;
}
int start = 0, end = A.length - 1;
int num = 0;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (A[mid] == target) {
num++;
int left, right;
left = mid - 1;
right = mid + 1;
while (left >= start && A[left] == A[mid]) {
num++;
left--;
}
while (right <= end && A[right] == A[mid]) {
num++;
right++;
}
return num;
} else if (A[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (A[start] == target) {
num++;
}
if (A[end] == target) {
num++;
}
return num;
}
}
2017-09-25
254.drop eggs
要求：提供两个鸡蛋，总共有n层楼，求出最坏情况下，需要扔多少次，才能找到恰好扔坏鸡蛋的楼层k
思路：转化成x + (x - 1) + ... + 1 >= n，x就是答案（我也不知道为什么）
其实还可以用动态规划来做
public class Solution {
/*
* @param n: An integer
* @return: The sum of a and b
*/
public int dropEggs(int n) {
// write your code here
long ans = 0;
for (int i = 1; ; i++) {
ans += (long)i;
if (ans >= (long)n) {
return i;
}
}
}
}
2017-09-25
460.找出k个最接近的数
要求：在升序数组中，找出与target最接近的k个数，返回结果时，从最接近到最远的数依次排序，若差值相等时先排小的数
思路：先找到最接近的数，然后在它两边再寻找k - 1个，用两个指针，判断两边谁的差值小谁先进
public class Solution {
/*
* @param A: an integer array
* @param target: An integer
* @param k: An integer
* @return: an integer array
*/
public int[] kClosestNumbers(int[] A, int target, int k) {
// write your code here
if (A == null || A.length == 0) {
return A;
}
if (k > A.length) {
return A;
}
int[] results = new int[k];
//find the first one
int index = findFirstOne(A, target);
int start = index;
int end = index + 1;
for (int i = 0; i < k; i++) {
//start < 0 , end >= length
if (start < 0) {
results[i] = A[end++];
} else if (end >= A.length) {
results[i] = A[start--];
} else {
//begin from the smaller
if (target - A[start] <= A[end] - target) {
results[i] = A[start--];
} else {
results[i] = A[end++];
}
}
}
return results;
}
private int findFirstOne(int[] A, int target) {
int start = 0, end = A.length - 1;
int diff = Math.abs(A[0] - target);
int index = 0;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (Math.abs(A[mid] - target) < diff) {
diff = Math.abs(A[mid] - target);
index = mid;
}
if (A[mid] == target) {
return mid;
} else if (A[mid] < target) {
start = mid;
} else {
end = mid;
}
}
if (Math.abs(A[start] - target) < diff) {
return start;
}
if (Math.abs(A[end] - target) < diff) {
return end;
}
return index;
}
}
2017-09-25
61.搜索区间
要求：找出target在数组中出现的区间
思路（自写）：找出第一个，然后从这开始循环判断
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
if (A == null || A.length == 0) {
return new int[]{-1, -1};
}
int[] result = new int[2];
int index = findFirstOne(A, target);
result[0] = index;
while (index != -1 && index + 1 < A.length
&& A[index] == A[index + 1]) {
index++;
}
result[1] = index;
return result;
}
private int findFirstOne(int[] A, int target) {
int start = 0, end = A.length - 1;
while (start + 1 < end) {
int mid = start + (end - start) / 2;
if (A[mid] >= target) {
end = mid;
} else {
start = mid;
}
}
if (A[start] == target) {
return start;
}
if (A[end] == target) {
return end;
}
return -1;
}
}
2017-09-27
3 - 二叉树与分治法
97.二叉树的最大深度
要求：如题
思路1：（答案）分治法，找出左右节点的最大值 + 1 = 当前父节点的深度
public int maxDepth(TreeNode root) {
if (root == null) {
return 0;
}
int left = maxDepth(root.left);
int right = maxDepth(root.right);
return Math.max(left, right) + 1;
}
2017-01-26
思路2：（答案）遍历法，需要全局变量depth，每次遍历到子节点curDepth + 1，更新全局变量depth，遍历到最后为止
public class Solution {
/**
* @param root: The root of binary tree.
* @return: An integer.
*/
private int depth; //全局变量
public int maxDepth(TreeNode root) {
depth = 0;
helper(root, 1);
return depth;
}
private void helper(TreeNode root, int curDepth) {
if (root == null) {
return;
}
if (curDepth > depth) {
depth = curDepth;
}
//遍历
helper(root.left, curDepth + 1);
helper(root.right, curDepth + 1);
}
}
2017-01-26
480.二叉树的所有路径
要求：如题
思路1：（答案）分治法，分成两条路径，然后把路径的节点都加入到返回的结果中
warn：除了判断root == null，还要判断叶节点，因为要加入叶它自己
public List<String> binaryTreePaths(TreeNode root) {
List<String> paths = new ArrayList<String>();
if (root == null) {
return paths;
}
List<String> leftPaths = binaryTreePaths(root.left);
List<String> rightPaths = binaryTreePaths(root.right);
for (String path : leftPaths) {
paths.add(root.val + "->" + path);
}
for (String path : rightPaths) {
paths.add(root.val + "->" + path);
}
return paths;
}
思路2：（答案）遍历法，paths作为全局变量，然后通过helper一步步构建每个path，仍然要判断叶
public List<String> binaryTreePaths(TreeNode root) {
List<String> paths = new ArrayList<String>(); //全局变量
if (root == null) {
return paths;
}
helper(root, root.val + "", paths);
return paths;
}
private void helper(TreeNode root, String path, List<String> paths) {
if (root == null) {
return;
}
//leaf
if (root.left == null && root.right == null) {
paths.add(path);
return;
}
if (root.left != null) {
helper(root.left, path + "->" + root.left.val, paths);
}
if (root.right != null) {
helper(root.right, path + "->" + root.right.val, paths);
}
}
2017-01-26
596.和最小的子树Minimum Subtree
要求：找出和最小的子数所对应的根节点
思路1：（答案）traverse + divide conquer，在helper中使用divide conquer的思想来算当前的和sum
warn：helper中返回值不能是void，因为算left 和 right要有值
public class Solution {
/**
* @param root the root of binary tree
* @return the root of the minimum subtree
*/
//全局变量
private TreeNode subtreeRoot = null;
private int subtreeSum = Integer.MAX_VALUE;
public TreeNode findSubtree(TreeNode root) {
//traverse
helper(root);
return subtreeRoot;
}
private int helper(TreeNode root) {
if (root == null) {
return 0;
}
//divide conquer
int sum = helper(root.left) + helper(root.right) + root.val;
if (subtreeSum > sum) {
subtreeSum = sum;
subtreeRoot = root;
}
return sum;
}
思路2：只用divide conquer，增加ResultType类来保存每次递归时的最小和minSum、对应的根节点subtreeRoot、当前的和Sum
public class Solution {
/*
* @param root: the root of binary tree
* @return: the root of the minimum subtree
*/
public class ResultType {
public TreeNode subtreeRoot;
public int minSum, sum;
public ResultType(TreeNode subtreeRoot, int minSum, int sum) {
this.subtreeRoot = subtreeRoot;
this.minSum = minSum;
this.sum = sum;
}
}
public TreeNode findSubtree(TreeNode root) {
// write your code here
ResultType result = dfsHelper(root);
return result.subtreeRoot;
}
private ResultType dfsHelper(TreeNode root) {
if (root == null) {
return new ResultType(null, Integer.MAX_VALUE, 0);
}
ResultType left = dfsHelper(root.left);
ResultType right = dfsHelper(root.right);
ResultType curr = new ResultType(root, left.sum + right.sum + root.val,
left.sum + right.sum + root.val);
if (left.minSum < curr.minSum) {
curr.minSum = left.minSum;
curr.subtreeRoot = left.subtreeRoot;
}
if (right.minSum < curr.minSum) {
curr.minSum = right.minSum;
curr.subtreeRoot = right.subtreeRoot;
}
return curr;
}
}
2017-10-12
2017-01-26
597.平均值最大的子树Subtree with Maximum Average
要求：找出平均值最大的子树对应的根节点
思路：（答案）traverse + divide conquer，加入ResultType类来保存sum和size（这里创建了ResultType类的全局变量，也就是体现了traverse的思想），在helper中先用分治思想算出当前的sum与size，更新较大average保存在全局变量中，helper同时也保存当前的ResultType，因为这样才会有left 和 right的数据，由下层的递归得到
warn：在对比average时，若通过sum除以size会带来误差，应当转成乘法的不等式
/**
* 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 {
//用ResultType来存储sum和size获得average
private class ResultType {
public int sum;
public int size;
public ResultType(int sum, int size) {
this.sum = sum;
this.size = size;
}
}
//全局变量
private TreeNode subtreeRoot = null;
private ResultType subtreeResult = null;
public TreeNode findSubtree2(TreeNode root) {
helper(root);
return subtreeRoot;
}
//有返回值的helper
private ResultType helper(TreeNode root) {
if (root == null) {
return new ResultType(0, 0);
}
//divide & conquer
ResultType left = helper(root.left);
ResultType right = helper(root.right);
//当前的结果
ResultType currResult = new ResultType(left.sum + right.sum + root.val,
left.size + right.size + 1);
//相乘防止误差
if (subtreeRoot == null || subtreeResult.sum * currResult.size
< currResult.sum * subtreeResult.size) {
subtreeResult = currResult;
subtreeRoot = root;
}
return currResult;
}
}
2017-01-27
453.二叉树的扁平化Flatten Binary Tree to Linked List
要求：如题
思路1：（答案）divide conquer，由于要返回最后一个点，所以要helper方法，把左子树最后一个连接右子树第一个，并把root的左节点代替右节点
warn：helper中作非空判断！
/**
* 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: a TreeNode, the root of the binary tree
* @return: nothing
*/
public void flatten(TreeNode root) {
helper(root);
}
private TreeNode helper(TreeNode root) {
if (root == null) {
return null;
}
//把左右子树分别扁平化并找出最后一个数
TreeNode leftLast = helper(root.left);
TreeNode rightLast = helper(root.right);
//左子树最后一个连接右子树第一个，并把root的左节点代替右节点
if (leftLast != null) {
leftLast.right = root.right;
root.right = root.left;
root.left = null;
}
//特殊判断
if (rightLast != null) {
return rightLast;
}
if (leftLast != null) {
return leftLast;
}
return root;
}
}
思路2：（答案）traverse，记录上一个点，然后把上一个点与当前进行连接
warn：要保存当前节点的右子节点？？
/**
* 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: a TreeNode, the root of the binary tree
* @return: nothing
*/
//全局变量表示上一个节点
private TreeNode lastNode = null;
public void flatten(TreeNode root) {
if (root == null) {
return;
}
//上个节点连接当前
if (lastNode != null) {
lastNode.left = null;
lastNode.right = root;
}
//更新lastNode
lastNode = root;
//保存当前节点的右子节点
TreeNode right = root.right;
flatten(root.left);
flatten(right);
}
}
思路3：还可以非递归来做，代码还没看懂
2017-01-27
595.二叉树的最长连续序列Binary Tree Longest Consecutive Sequence
要求：只能从上到下
思路1：（看完思路自写AC）traverse + divide conquer，helper中返回int类型
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
/**
* @param root the root of binary tree
* @return the length of the longest consecutive sequence path
*/
private int subtreeLen;
public int longestConsecutive(TreeNode root) {
subtreeLen = 0;
helper(root);
return subtreeLen;
}
private int helper(TreeNode root) {
if (root == null) {
return 0;
}
int left = helper(root.left);
int right = helper(root.right);
int currLen = 1; //最起码有根一个
if (root.left != null && root.val + 1 == root.left.val) {
currLen = left + 1;
}
if (root.right != null && root.val + 1 == root.right.val) {
currLen = right + 1;
}
if (currLen > subtreeLen) {
subtreeLen = currLen;
}
return currLen;
}
}
思路2：（根据答案自写）ResultType + divide conquer，ResultType包括maxFromRoot和maxInSubtree
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
/**
* @param root the root of binary tree
* @return the length of the longest consecutive sequence path
*/
private class ResultType{
public int maxFromRoot;
public int maxInSubtree;
public ResultType(int maxFromRoot, int maxInSubtree) {
this.maxFromRoot = maxFromRoot;
this.maxInSubtree = maxInSubtree;
}
}
public int longestConsecutive(TreeNode root) {
return helper(root).maxInSubtree;
}
private ResultType helper(TreeNode root) {
if (root == null) {
return new ResultType(0, 0);
}
ResultType left = helper(root.left);
ResultType right = helper(root.right);
ResultType result = new ResultType(1, 0);
if (root.left != null && root.val + 1 == root.left.val) {
result.maxFromRoot = left.maxFromRoot + 1;
}
if (root.right != null && root.val + 1 == root.right.val) {
result.maxFromRoot = right.maxFromRoot + 1;
}
result.maxInSubtree = Math.max(result.maxFromRoot,
Math.max(left.maxInSubtree, right.maxInSubtree));
return result;
}
}
2017-01-30
376.二叉树的路径和Binary tree Path Sum
要求：找出所有从上到下的和等于target的路径
思路：（看思路自写AC）traverse，用子集的方法，然后判断if (sum == target)，把当前path加入到paths
/**
* 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
* @param target an integer
* @return all valid paths
*/
public List<List<Integer>> binaryTreePathSum(TreeNode root, int target) {
List<List<Integer>> paths = new ArrayList<>();
if (root == null) {
return paths;
}
//先把根放进去
List<Integer> path = new ArrayList<>();
path.add(root.val);
helper(root, root.val, target, path, paths);
return paths;
}
private void helper(TreeNode root, int sum, int target, List<Integer> path,
List<List<Integer>> paths) {
if (root.left == null && root.right == null) {
if (sum == target) {
paths.add(new ArrayList<Integer>(path));
}
return;
}
if (root.left != null) {
path.add(root.left.val);
helper(root.left, sum + root.left.val, target, path, paths);
path.remove(path.size() - 1);
}
if (root.right != null) {
path.add(root.right.val);
helper(root.right, sum + root.right.val, target, path, paths);
path.remove(path.size() - 1);
}
}
}
2017-01-30
93.平衡二叉树Balanced Binary Tree
要求：判断左右子树的深度是否相差1以内
思路：（答案）divide conquer + ResultType，helper返回的是isBalanced和maxDepth
warn：递归中不仅要判断当前根节点是否平衡，还要判断它的两个子节点是否平衡：if (! left.isBalanced || ! right.isBalanced)，只要两边有一个不平衡就整个不平衡
2017-10-12
/**
* 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: True if this Binary tree is Balanced, or false.
*/
private class ResultType {
public boolean isBalanced;
public int maxDepth;
public ResultType(boolean isBalanced, int maxDepth) {
this.isBalanced = isBalanced;
this.maxDepth = maxDepth;
}
}
public boolean isBalanced(TreeNode root) {
return helper(root).isBalanced;
}
private ResultType helper(TreeNode root) {
if (root == null) {
return new ResultType(true, 0);
}
ResultType left = helper(root.left);
ResultType right = helper(root.right);
if (Math.abs(left.maxDepth - right.maxDepth) > 1) {
return new ResultType(false, -1);
}
if (!left.isBalanced || !right.isBalanced) {
return new ResultType(false, -1);
}
int currDepth = Math.max(left.maxDepth, right.maxDepth) + 1;
return new ResultType(true, currDepth);
}
}
2017-01-30
4 - 宽度优先搜索
69.二叉树的层级遍历
要求：如题
思路：（自写但没有bugfree）就是bfs的模板，分层表示要加size = queue.size()
/**
* 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: Level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> results = new ArrayList<>();
if (root == null) {
return results;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
//bfs
while (!queue.isEmpty()) {
ArrayList<Integer> level = new ArrayList<>();
//分层显示
int size = queue.size();
for (int i = 0; i < size; i++) {
//保存头节点
TreeNode head = queue.poll();
level.add(head.val);
if (head.left != null) {
queue.offer(head.left);
}
if (head.right != null) {
queue.offer(head.right);
}
}
results.add(level);
}
return results;
}
}
2017-02-06
242.将二叉树按照层级转化成链表
要求：如题
思路：自写AC 与69类似，加入了链表的基本操作
public class Solution {
/**
* @param root the root of binary tree
* @return a lists of linked list
*/
public List<ListNode> binaryTreeToLists(TreeNode root) {
// Write your code here
List<ListNode> results = new ArrayList<>();
if (root == null) {
return results;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
while (!queue.isEmpty()) {
ListNode dummy = new ListNode(0);
ListNode node = dummy;
int size = queue.size();
for (int i = 0; i < size; i++) {
TreeNode head = queue.poll();
ListNode curr = new ListNode(head.val);
node.next = curr;
node = node.next;
if (head.left != null) {
queue.offer(head.left);
}
if (head.right != null) {
queue.offer(head.right);
}
}
results.add(dummy.next);
}
return results;
}
}
2017-10-13
7.二叉树的序列化和反序列化
要求：如题
思路：序列化就直接bfs且加入时包括中间出现的null（#）并另外去除尾部的null，反序列化的时候需要注意判断左右儿子和当前在操作哪个节点
/**
* 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;
* }
* }
*/
class Solution {
/**
* This method will be invoked first, you should design your own algorithm
* to serialize a binary tree which denote by a root node to a string which
* can be easily deserialized by your own "deserialize" method later.
*/
public String serialize(TreeNode root) {
if (root == null) {
return "{}";
}
ArrayList<TreeNode> list = new ArrayList<>();
list.add(root);
//bfs
for (int i = 0; i < list.size(); i++) {
TreeNode node = list.get(i);
if (node == null) {
continue;
}
list.add(node.left);
list.add(node.right);
}
while (list.get(list.size() - 1) == null) {
list.remove(list.size() - 1);
}
StringBuilder res = new StringBuilder();
//把头节点加入
res.append("{" + root.val);
for (int i = 1; i < list.size(); i++) {
if (list.get(i) != null) {
res.append("," + list.get(i).val);
} else {
res.append(",#");
}
}
res.append("}");
return res.toString();
} /**
* This method will be invoked second, the argument data is what exactly
* you serialized at method "serialize", that means the data is not given by
* system, it's given by your own serialize method. So the format of data is
* designed by yourself, and deserialize it here as you serialize it in
* "serialize" method.
*/
public TreeNode deserialize(String data) {
if (data.equals("{}")) {
return null;
}
//排除, {}并拆分
String[] val = data.substring(1, data.length() - 1).split(",");
ArrayList<TreeNode> list = new ArrayList<>();
TreeNode root = new TreeNode(Integer.parseInt(val[0]));
list.add(root);
//指针标识当前操作哪个节点，以及判断是左还是右儿子
int index = 0;
boolean isLeftChild = true;
for (int i = 1; i < val.length; i++) {
if (!val[i].equals("#")) {
TreeNode node = new TreeNode(Integer.parseInt(val[i]));
//非空的才放进去list中
list.add(node);
if (isLeftChild) {
list.get(index).left = node;
} else {
list.get(index).right = node;
}
}
//到右儿子就操作下一个节点
if (!isLeftChild) {
index++;
}
//操作完左儿子变右儿子
isLeftChild = !isLeftChild;
}
return root;
}
}
2017-02-06
178.图是否是树
要求：如题，图用二维数组（邻接表）表示，二维数组有两列m（边数）行，数组的每个点就是图的点，同一行的点代表是neighbor
思路：图是树有两个条件：1.边数为点数 - 1；2.能连通
warn：构造无向图时一定要把双向都加入，所以在bfs时用另一个set去除已经加入过queue的neighbor
public class Solution {
/**
* @param n an integer
* @param edges a list of undirected edges
* @return true if it's a valid tree, or false
*/
public boolean validTree(int n, int[][] edges) {
if (n == 0) {
return false;
}
//1.树的边数应为点数 - 1
if (edges.length != n - 1) {
return false;
}
//用邻接表定义图
Map<Integer, Set<Integer>> graph = initializeGraph(n, edges);
//2.从一个点出发能跑通
Queue<Integer> queue = new LinkedList<>();
Set<Integer> set = new HashSet<>();
queue.offer(0);
//排除相同点进入队列
set.add(0);
int visited = 0;
while (!queue.isEmpty()) {
int node = queue.poll();
visited++;
for (int neighbor : graph.get(node)) {
if (set.contains(neighbor)) {
continue;
}
queue.offer(neighbor);
set.add(neighbor);
}
}
return visited == n;
}
private Map<Integer, Set<Integer>> initializeGraph(int n, int[][] edges) {
//初始化HashMap
Map<Integer, Set<Integer>> graph = new HashMap<>();
for (int i = 0; i < n; i++) {
graph.put(i, new HashSet<Integer>());
}
//把边放入
for (int i = 0; i < edges.length; i++) {
int u = edges[i][0];
int v = edges[i][1];
graph.get(u).add(v);
graph.get(v).add(u);
}
return graph;
}
}
2017-02-07
137.克隆图
要求：把图的点和边都复制
思路：1.bfs所有点，放进ArrayList中（相当于获得所有节点的引用）2.deep copy，真正new一个新的一样的节点，并把新老节点用map对应 3.运用老节点帮新节点找到邻居
warn：理解deep copy，分清楚创建的是新对象还是新引用
/**
* Definition for undirected graph.
* class UndirectedGraphNode {
* int label;
* ArrayList<UndirectedGraphNode> neighbors;
* UndirectedGraphNode(int x) { label = x; neighbors = new ArrayList<UndirectedGraphNode>(); }
* };
*/
public class Solution {
/**
* @param node: A undirected graph node
* @return: A undirected graph node
*/
public UndirectedGraphNode cloneGraph(UndirectedGraphNode node) {
if (node == null) {
return null;
}
//bfs来获得所有点
ArrayList<UndirectedGraphNode> nodes = getNodes(node);
//deep copy，并建立新老node的对应关系
Map<UndirectedGraphNode, UndirectedGraphNode> map = new HashMap<>();
for (UndirectedGraphNode n : nodes) {
map.put(n, new UndirectedGraphNode(n.label));
}
//copy neighbor
for (UndirectedGraphNode n : nodes) {
UndirectedGraphNode newNode = map.get(n);
for (UndirectedGraphNode neighbor : n.neighbors) {
UndirectedGraphNode newNeighbor = map.get(neighbor);
newNode.neighbors.add(newNeighbor);
}
}
return map.get(node);
}
//标准对于图的bfs
private ArrayList<UndirectedGraphNode> getNodes(UndirectedGraphNode node) {
Queue<UndirectedGraphNode> queue = new LinkedList<>();
Set<UndirectedGraphNode> set = new HashSet<>();
queue.offer(node);
set.add(node);
while (!queue.isEmpty()) {
UndirectedGraphNode head = queue.poll();
for (UndirectedGraphNode neighbor : head.neighbors) {
if (!set.contains(neighbor)) {
set.add(neighbor);
queue.offer(neighbor);
}
}
}
//把set deep copy到ArrayList中
return new ArrayList<UndirectedGraphNode>(set);
}
}
2017-02-07
618.Search Graph Nodes
要求：找出value为target的最近的点
思路：bfs找出满足条件即可
/**
* Definition for graph node.
* class UndirectedGraphNode {
* int label;
* ArrayList<UndirectedGraphNode> neighbors;
* UndirectedGraphNode(int x) {
* label = x; neighbors = new ArrayList<UndirectedGraphNode>();
* }
* };
*/
public class Solution {
/**
* @param graph a list of Undirected graph node
* @param values a hash mapping, <UndirectedGraphNode, (int)value>
* @param node an Undirected graph node
* @param target an integer
* @return the a node
*/
public UndirectedGraphNode searchNode(ArrayList<UndirectedGraphNode> graph,
Map<UndirectedGraphNode, Integer> values,
UndirectedGraphNode node,
int target) {
Queue<UndirectedGraphNode> queue = new LinkedList<>();
Set<UndirectedGraphNode> set = new HashSet<>();
queue.offer(node);
set.add(node);
while (!queue.isEmpty()) {
UndirectedGraphNode head = queue.poll();
if (values.get(head) == target) {
return head;
}
for (UndirectedGraphNode neighbor : head.neighbors) {
if (!set.contains(neighbor)) {
set.add(neighbor);
queue.offer(neighbor);
}
}
}
return null;
}
}
2017-02-08
433.岛屿的个数
要求：矩阵中1代表岛屿，0代表海洋，相邻的1代表同一个岛，求岛的个数
思路：用bfs把1附近的1都标记为0
warn：判断越界，写inBound函数
public class Solution {
/**
* @param grid a boolean 2D matrix
* @return an integer
*/
class Node {
int x;
int y;
public Node(int x, int y) {
this.x = x;
this.y = y;
}
}
public int numIslands(boolean[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int row = grid.length;
int col = grid[0].length;
int island = 0;
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if (grid[i][j]) {
isNeighbor(grid, i, j);
island++;
}
}
}
return island;
}
private void isNeighbor(boolean[][] grid, int x, int y) {
//坐标变换数组
int[] deltaX = {0, 1, -1, 0};
int[] deltaY = {1, 0, 0, -1};
Queue<Node> queue = new LinkedList<>();
queue.offer(new Node(x, y));
//遍历过的标记为false
grid[x][y] = false;
//bfs访问四个方向
while (!queue.isEmpty()) {
Node head = queue.poll();
for (int i = 0; i < 4; i++) {
Node neighbor = new Node(head.x + deltaX[i], head.y + deltaY[i]);
//判断越界
if (!inBound(neighbor, grid)) {
continue;
}
if (grid[neighbor.x][neighbor.y]) {
grid[neighbor.x][neighbor.y] = false;
queue.offer(neighbor);
}
}
}
}
private boolean inBound(Node node, boolean[][] grid) {
int row = grid.length;
int col = grid[0].length;
return node.x >= 0 && node.x < row && node.y >= 0 && node.y < col;
}
}
2017-02-08
二刷 bfs时忘记把一开始的head放进去后，把它变成false，while循环中忘记每次更新head
2017-10-13
598.Zombie in Matrix
要求：0代表人，1代表僵尸，2代表墙，求多少天能够把人都变成僵尸
思路：1.数人头和把僵尸放到队列中 2.完整的三层循环的分层bfs（背），每一层就是经过一天 3.人数为0时返回天数
public class Solution {
/**
* @param grid a 2D integer grid
* @return an integer
*/
class Node {
int x, y;
public Node(int x, int y) {
this.x = x;
this.y = y;
}
}
public int PEOPLE = 0;
public int ZOMBIE = 1;
public int[] deltaX = {0, 1, -1, 0};
public int[] deltaY = {1, 0, 0, -1};
public int zombie(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return 0;
}
int people = 0;
Queue<Node> queue = new LinkedList<>();
int row = grid.length;
int col = grid[0].length;
//计算人的数量并把一开始的僵尸位置放到队列中
for (int i = 0; i < row; i++) {
for (int j = 0; j < col; j++) {
if (grid[i][j] == PEOPLE) {
people++;
} else if (grid[i][j] == ZOMBIE) {
queue.offer(new Node(i, j));
}
}
}
if (people == 0) {
return 0;
}
int days = 0;
//完整的bfs
while (!queue.isEmpty()) {
//分层
int size = queue.size();
days++;
for (int i = 0; i < size; i++) {
Node head = queue.poll();
for (int j = 0; j < 4; j++) {
Node neighbor = new Node(head.x + deltaX[j], head.y + deltaY[j]);
//判断是否越界以及是否为人
if (!isPeople(neighbor, grid)) {
continue;
}
grid[neighbor.x][neighbor.y] = ZOMBIE;
people--;
if (people == 0) {
return days;
}
queue.offer(neighbor);
}
}
}
return -1;
}
private boolean isPeople(Node node, int[][] grid) {
int row = grid.length;
int col = grid[0].length;
if (node.x < 0 || node.x >= row) {
return false;
}
if (node.y < 0 || node.y >= col) {
return false;
}
return grid[node.x][node.y] == PEOPLE;
}
}
2017-02-08
611.Knight Shortest Path
要求：骑士跳日字，计算从source位置到destination位置的步数
思路：（自写AC）1.坐标变换数组 2.完整的三层bfs 3.inBound方法
warn：不能用set标记已走过的地方，会超出内存范围，所以走过的地方标为BARRIER
/**
* Definition for a point.
* public class Point {
* public int x, y;
* public Point() { x = 0; y = 0; }
* public Point(int a, int b) { x = a; y = b; }
* }
*/
public class Solution {
/**
* @param grid a chessboard included 0 (false) and 1 (true)
* @param source, destination a point
* @return the shortest path
*/
public boolean EMPTY = false;
public boolean BARRIER = true;
public int[] deltaX = {1, 1, 2, 2, -1, -1, -2, -2};
public int[] deltaY = {2, -2, 1, -1, 2, -2, 1, -1};
public int shortestPath(boolean[][] grid, Point source, Point destination) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return -1;
}
if (source.x == destination.x && source.y == destination.y) {
return 0;
}
int rows = grid.length;
int cols = grid[0].length;
Queue<Point> queue = new LinkedList<>();
queue.offer(source);
//标记已走过
grid[source.x][source.y] = BARRIER;
//bfs
int length = 0;
while (!queue.isEmpty()) {
length++;
//分层
int size = queue.size();
for (int i = 0; i < size; i++) {
Point head = queue.poll();
//八个方向跳
for (int j = 0; j < 8; j++) {
Point next = new Point(head.x + deltaX[j], head.y + deltaY[j]);
if (next.x == destination.x && next.y == destination.y) {
return length;
}
if (isBarrier(next, grid)) {
continue;
}
queue.offer(next);
grid[next.x][next.y] = BARRIER;
}
}
}
return -1;
}
private boolean isBarrier(Point p, boolean[][] grid) {
int rows = grid.length;
int cols = grid[0].length;
if (p.x < 0 || p.x >= rows) {
return true;
}
if (p.y < 0 || p.y >= cols) {
return true;
}
return grid[p.x][p.y] == BARRIER;
}
}
2017-02-09
573.Build Post Office II
要求：0代表office能建的位置，1代表房，2代表墙，求office能够到达所有房的最短路径
思路：求每间房到所有0点的距离，当求每个点时就可以把对应房间的距离加起来
class Coordinate {
int x, y;
public Coordinate(int x, int y) {
this.x = x;
this.y = y;
}
}
public class Solution {
public int EMPTY = 0;
public int HOUSE = 1;
public int WALL = 2;
public int[][] grid;
public int n, m;
public int[] deltaX = {0, 1, -1, 0};
public int[] deltaY = {1, 0, 0, -1};
private List<Coordinate> getCoordinates(int type) {
List<Coordinate> coordinates = new ArrayList<>();
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == type) {
coordinates.add(new Coordinate(i, j));
}
}
}
return coordinates;
}
private void setGrid(int[][] grid) {
n = grid.length;
m = grid[0].length;
this.grid = grid;
}
private boolean inBound(Coordinate coor) {
if (coor.x < 0 || coor.x >= n) {
return false;
}
if (coor.y < 0 || coor.y >= m) {
return false;
}
return grid[coor.x][coor.y] == EMPTY;
}
/**
* @param grid a 2D grid
* @return an integer
*/
public int shortestDistance(int[][] grid) {
if (grid == null || grid.length == 0 || grid[0].length == 0) {
return -1;
}
// set n, m, grid
setGrid(grid);
List<Coordinate> houses = getCoordinates(HOUSE);
int[][] distanceSum = new int[n][m];;
int[][] visitedTimes = new int[n][m];;
for (Coordinate house : houses) {
bfs(house, distanceSum, visitedTimes);
}
int shortest = Integer.MAX_VALUE;
List<Coordinate> empties = getCoordinates(EMPTY);
for (Coordinate empty : empties) {
if (visitedTimes[empty.x][empty.y] != houses.size()) {
continue;
}
shortest = Math.min(shortest, distanceSum[empty.x][empty.y]);
}
if (shortest == Integer.MAX_VALUE) {
return -1;
}
return shortest;
}
private void bfs(Coordinate start,
int[][] distanceSum,
int[][] visitedTimes) {
Queue<Coordinate> queue = new LinkedList<>();
boolean[][] hash = new boolean[n][m];
queue.offer(start);
hash[start.x][start.y] = true;
int steps = 0;
while (!queue.isEmpty()) {
steps++;
int size = queue.size();
for (int temp = 0; temp < size; temp++) {
Coordinate coor = queue.poll();
for (int i = 0; i < 4; i++) {
Coordinate adj = new Coordinate(
coor.x + deltaX[i],
coor.y + deltaY[i]
);
if (!inBound(adj)) {
continue;
}
if (hash[adj.x][adj.y]) {
continue;
}
queue.offer(adj);
hash[adj.x][adj.y] = true;
distanceSum[adj.x][adj.y] += steps;
visitedTimes[adj.x][adj.y]++;
} // direction
} // for temp
} // while
}
}
（答案代码未仔细看，类似于上两题）
2017-02-09
127.拓扑排序
要求：如题，从入度0的出发，找出其中一个解
思路：（自写AC）1. 算每个点的入度 2. bfs每个点，对应的neighbor的入度 - 1，到入度为0时进入队列
/**
* 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> order = new ArrayList<>();
if (graph == null) {
return order;
}
// 1. count indegree
Map<DirectedGraphNode, Integer> indegree = getIndegree(graph);
// 2. bfs
Queue<DirectedGraphNode> queue = new LinkedList<>();
// find first node
for (DirectedGraphNode node : graph) {
if (indegree.get(node) == 0) {
queue.offer(node);
order.add(node);
}
}
while (!queue.isEmpty()) {
DirectedGraphNode head = queue.poll();
for (DirectedGraphNode neighbor : head.neighbors) {
indegree.put(neighbor, indegree.get(neighbor) - 1);
if (indegree.get(neighbor) == 0) {
queue.offer(neighbor);
order.add(neighbor);
}
}
}
return order;
}
private Map<DirectedGraphNode, Integer> getIndegree(ArrayList<DirectedGraphNode> graph) {
Map<DirectedGraphNode, Integer> map = new HashMap<>();
// 初始化
for (DirectedGraphNode node : graph) {
map.put(node, 0);
}
for (DirectedGraphNode node : graph) {
for (DirectedGraphNode neighbor : node.neighbors) {
map.put(neighbor, map.get(neighbor) + 1);
}
}
return map;
}
}
2017-02-09
616.Course Schedule II
要求：把课程拓扑排序，给出的是二维数组表示的课程之间的关系
思路：（自写AC！） 1. 生成图 2. 算入度 3.bfs（与普通拓扑类似）
warn：注意特殊情况，判断是否有解！（index == numCourses时有解）
自写：
public class Solution {
/**
* @param numCourses a total of n courses
* @param prerequisites a list of prerequisite pairs
* @return the course order
*/
public int[] findOrder(int numCourses, int[][] prerequisites) {
int[] order = new int[numCourses];
if (numCourses == 0) {
return order;
}
//exceptional case
if (numCourses != 0 && (prerequisites == null
|| prerequisites.length == 0 || prerequisites[0].length == 0)) {
for (int i = 0; i < numCourses; i++) {
order[i] = i;
}
return order;
}
// 1. initialize graph
Map<Integer, ArrayList<Integer>> graph = initializeGraph(numCourses, prerequisites);
// 2. count indegree
Map<Integer, Integer> indegree = getIndegree(graph, numCourses);
// 3. bfs
int index = 0;
Queue<Integer> queue = new LinkedList<>();
// find start
for (int i = 0; i < numCourses; i++) {
if (indegree.get(i) == 0) {
queue.offer(i);
order[index++] = i;
}
}
while (!queue.isEmpty()) {
int head = queue.poll();
for (int neighbor : graph.get(head)) {
indegree.put(neighbor, indegree.get(neighbor) - 1);
if (indegree.get(neighbor) == 0) {
queue.offer(neighbor);
order[index++] = neighbor;
}
}
}
if (index == numCourses) {
return order;
}
return new int[0];
}
private Map<Integer, ArrayList<Integer>> initializeGraph(int numCourses, int[][] prerequisites) {
Map<Integer, ArrayList<Integer>> graph = new HashMap<>();
for (int i = 0; i < numCourses; i++) {
graph.put(i, new ArrayList<Integer>());
}
for (int i = 0; i < prerequisites.length; i++) {
int u = prerequisites[i][0];
int v = prerequisites[i][1];
graph.get(v).add(u);
}
return graph;
}
private Map<Integer, Integer> getIndegree(Map<Integer, ArrayList<Integer>> graph, int numCourses) {
Map<Integer, Integer> indegree = new HashMap<>();
for (int i = 0; i < numCourses; i++) {
indegree.put(i, 0);
}
for (int i = 0; i < numCourses; i++) {
for (Integer neighbor : graph.get(i)) {
indegree.put(neighbor, indegree.get(neighbor) + 1);
}
}
return indegree;
}
}
答案写法：
public class Solution {
/**
* @param numCourses a total of n courses
* @param prerequisites a list of prerequisite pairs
* @return the course order
*/
public int[] findOrder(int numCourses, int[][] prerequisites) {
// Write your code here
List[] edges = new ArrayList[numCourses];
int[] degree = new int[numCourses];
for (int i = 0;i < numCourses; i++)
edges[i] = new ArrayList<Integer>();
for (int i = 0; i < prerequisites.length; i++) {
degree[prerequisites[i][0]] ++ ;
edges[prerequisites[i][1]].add(prerequisites[i][0]);
}
Queue queue = new LinkedList();
for(int i = 0; i < degree.length; i++){
if (degree[i] == 0) {
queue.add(i);
}
}
int count = 0;
int[] order = new int[numCourses];
while(!queue.isEmpty()){
int course = (int)queue.poll();
order[count] = course;
count ++;
int n = edges[course].size();
for(int i = n - 1; i >= 0 ; i--){
int pointer = (int)edges[course].get(i);
degree[pointer]--;
if (degree[pointer] == 0) {
queue.add(pointer);
}
}
}
if (count == numCourses)
return order;
return new int[0];
}
}
2017-02-09
还缺一道hard题
5 - 深度优先搜索
18.subsets II
要求：集合中有重复元素
思路：dfs，在I的前提下加入判断条件i != 0 && nums[i] == nums[i - 1] && i != startIndex时continue
public ArrayList<ArrayList<Integer>> subsetsWithDup(int[] nums) {
ArrayList<ArrayList<Integer>> results = new ArrayList<>();
if (nums == null || nums.length == 0) {
return results;
}
Arrays.sort(nums);
dfsHelper(nums, 0, new ArrayList<Integer>(), results);
return results;
}
private void dfsHelper(int[] nums, int startIndex, ArrayList<Integer> subset,
ArrayList<ArrayList<Integer>> results) {
//deep copy
results.add(new ArrayList<Integer>(subset));
for (int i = startIndex; i < nums.length; i++) {
if (i != 0 && nums[i] == nums[i - 1] && i > startIndex) {
continue;
}
subset.add(nums[i]);
dfsHelper(nums, i + 1, subset, results);
subset.remove(subset.size() - 1);
}
}
2017-02-10
135.数字组合 combination sum
要求：组合中可能存在重复数，每个数能用多次，求出能相加等于target的集合
思路：dfs，注意条件判断就好，removeDuplicate方法
public class Solution {
/**
* @param candidates: A list of integers
* @param target:An integer
* @return: A list of lists of integers
*/
public List<List<Integer>> combinationSum(int[] candidates, int target) {
List<List<Integer>> results = new ArrayList<>();
if (candidates == null || candidates.length == 0) {
return results;
}
int[] nums = removeDuplicates(candidates);
dfsHelper(nums, target, 0, new ArrayList<Integer>(), results);
return results;
}
private void dfsHelper(int[] nums, int target, int startIndex,
List<Integer> subset,
List<List<Integer>> results) {
if (target == 0) {
results.add(new ArrayList<Integer>(subset));
}
for (int i = startIndex; i < nums.length; i++) {
if (nums[i] > target) {
break;
}
subset.add(nums[i]);
dfsHelper(nums, target - nums[i], i, subset, results);
subset.remove(subset.size() - 1);
}
}
private int[] removeDuplicates(int[] candidates) {
Arrays.sort(candidates);
int index = 0;
for (int i = 1; i < candidates.length; i++) {
if (candidates[i] != candidates[index]) {
candidates[++index] = candidates[i];
}
}
int[] nums = new int[index + 1];
for (int i = 0; i < index + 1; i++) {
nums[i] = candidates[i];
}
return nums;
}
}
2017-02-10
153.数字组合 combination sum II
要求：组合中有重复数，每个数只能用一次
思路：与上题区别是这题不能直接去重，因为相同的数也要加入到组合中，但需要像subset II那样排除相同的数调换顺序的重复情况
public List<List<Integer>> combinationSum2(int[] num, int target) {
List<List<Integer>> combinations = new ArrayList<>();
if (num == null || num.length == 0) {
return combinations;
}
Arrays.sort(num);
dfsHelper(num, target, 0, new ArrayList<Integer>(), combinations);
return combinations;
}
private void dfsHelper(int[] num, int target, int startIndex,
List<Integer> combination,
List<List<Integer>> combinations) {
if (target == 0) {
combinations.add(new ArrayList<Integer>(combination));
return;
}
for (int i = startIndex; i < num.length; i++) {
if (i != 0 && num[i] == num[i - 1] && i != startIndex) {
continue;
}
if (target < num[i]) {
break;
}
combination.add(num[i]);
dfsHelper(num, target - num[i], i + 1, combination, combinations);
combination.remove(combination.size() - 1);
}
}
2017-02-10
136.分隔回文串 Palindrome Partitioning
要求：找出所有把字符串s切割成的全部子字符串都是回文串的方法
思路：仍然是dfs模板，每次加入的是从startIndex切割到当前i的子字符串，并判断其是否回文串（加入isPalindrome方法），递归的出口是切完（startIndex == s.length()）
public List<List<String>> partition(String s) {
List<List<String>> results = new ArrayList<>();
if (s == null || s.length() == 0) {
return results;
}
dfsHelper(s, 0, new ArrayList<String>(), results);
return results;
}
private void dfsHelper(String s, int startIndex,
List<String> partition,
List<List<String>> results) {
//切完了
if (startIndex == s.length()) {
results.add(new ArrayList<String>(partition));
return;
}
for (int i = startIndex; i < s.length(); i++) {
//切从startIndex到i的一段
String ss = s.substring(startIndex, i + 1);
if (!isPalindrome(ss)) {
continue;
}
partition.add(ss);
dfsHelper(s, i + 1, partition, results);
partition.remove(partition.size() - 1);
}
}
private 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;
}
2017-02-10
15.全排序
要求：如题
思路：dfs，用visited数组标记是否访问过
warn：其实nums == null和nums.length == 0不一样
class Solution {
/**
* @param nums: A list of integers.
* @return: A list of permutations.
*/
public List<List<Integer>> permute(int[] nums) {
List<List<Integer>> lists = new ArrayList<>();
if (nums == null) {
return lists;
}
if (nums.length == 0) {
lists.add(new ArrayList<Integer>());
return lists;
}
//sign if it has been visited
int[] visited = new int[nums.length];
helper(nums, visited, new ArrayList<Integer>(), lists);
return lists;
}
private void helper(int[] nums, int [] visited,
List<Integer> list, List<List<Integer>> lists) {
if (list.size() == nums.length) {
lists.add(new ArrayList<Integer>(list));
}
for (int i = 0; i < nums.length; i++) {
if (visited[i] == 1) {
continue;
}
visited[i] = 1;
list.add(nums[i]);
helper(nums, visited, list, lists);
list.remove(list.size() - 1);
visited[i] = 0;
}
}
}
2017-02-13
16.带重复元素的排列
要求：如题
思路：与上题一样，并加入了subset II的条件
tips：判断时要加入visited[i - 1] == 0才continue
warn：别忘记排序！！
class Solution {
/**
* @param nums: A list of integers.
* @return: A list of unique permutations.
*/
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> lists = new ArrayList<>();
List<Integer> list = new ArrayList<>();
if (nums == null) {
return lists;
}
if (nums.length == 0) {
lists.add(list);
return lists;
}
Arrays.sort(nums);
int[] visited = new int[nums.length];
helper(nums, visited, list, lists);
return lists;
}
private void helper(int[] nums, int[] visited,
List<Integer> list, List<List<Integer>> lists) {
if (list.size() == nums.length) {
lists.add(new ArrayList<Integer>(list));
}
for (int i = 0; i < nums.length; i++) {
if (visited[i] == 1) {
continue;
}
if (i != 0 && nums[i] == nums[i - 1] && visited[i - 1] == 0) {
continue;
}
visited[i] = 1;
list.add(nums[i]);
helper(nums, visited, list, lists);
list.remove(list.size() - 1);
visited[i] = 0;
}
}
}
2017-02-13
6 - 链表与数组
604.womdow sum
要求：把k长度窗口里的和都算出
思路：（自写AC）two pointers
warn：边界判断
public int[] winSum(int[] nums, int k) {
if (nums == null || k == 0 || nums.length < k) {
return new int[0];
}
int[] sums = new int[nums.length - k + 1];
int sum = 0;
for (int index = 0; index < k; index++) {
sum += nums[index];
}
sums[0] = sum;
int i = 0;
int j = k - 1;
while (j < nums.length - 1) {
sum = sum - nums[i++] + nums[++j];
sums[i] = sum;
}
return sums;
}
2017-02-14
138.subarray sum
要求：找出和为0的子数组
思路：（自写AC）prefixSum，找出前缀和相同的项就代表此段子数组为0
写法1：双重循环找：
public ArrayList<Integer> subarraySum(int[] nums) {
ArrayList<Integer> result = new ArrayList<Integer>();
if (nums == null || nums.length == 0) {
return result;
}
int[] prefixSum = new int[nums.length];
int sum = 0;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
prefixSum[i] = sum;
}
if (prefixSum[nums.length - 1] == 0) {
result.add(0);
result.add(nums.length - 1);
return result;
}
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
if (prefixSum[i] == prefixSum[j]) {
result.add(i + 1);
result.add(j);
return result;
}
}
}
return result;
}
写法2：HashMap找（答案）
public ArrayList<Integer> subarraySum(int[] nums) {
// write your code here
int len = nums.length;
ArrayList<Integer> ans = new ArrayList<Integer>();
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
map.put(0, -1);
int sum = 0;
for (int i = 0; i < len; i++) {
sum += nums[i];
if (map.containsKey(sum)) {
ans.add(map.get(sum) + 1);
ans.add(i);
return ans;
}
map.put(sum, i);
}
return ans;
}
写法3：二分查找（略）
2017-02-14
41.maximum subarray
要求：找出最大子数组
思路：（自写一次AC）prefixSum数组
warn：注意index从0开始时的情况
public int maxSubArray(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
if (nums.length == 1) {
return nums[0];
}
//prefixSum
int sum = 0;
int[] prefixSum = new int[nums.length];
int result = Integer.MIN_VALUE;
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
prefixSum[i] = sum;
result = Math.max(prefixSum[i], result);
}
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
result = Math.max(prefixSum[j] - prefixSum[i], result);
}
}
return result;
}
2017-02-14
139.subarray sum cloest
要求：找出和最接近0的子数组
思路：（自写）prefixSum，每次更新一次，但用两次for循环的话数据量大时会TLE
public int[] subarraySumClosest(int[] nums) {
if (nums == null || nums.length == 0) {
return new int[0];
}
int[] res = new int[2];
int subarraySum = Integer.MAX_VALUE;
int sum = 0;
int[] prefixSum = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
sum += nums[i];
prefixSum[i] = sum;
if (subarraySum > Math.abs(prefixSum[i])) {
subarraySum = Math.abs(prefixSum[i]);
res[0] = 0;
res[1] = i;
}
}
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
int currSum = Math.abs(prefixSum[j] - prefixSum[i]);
if (subarraySum > currSum) {
subarraySum = currSum;
res[0] = i + 1;
res[1] = j;
}
}
}
return res;
}
答案（优化写法）：
class Pair {
int sum;
int index;
public Pair(int s, int i) {
sum = s;
index = i;
}
}
public class Solution {
/**
* @param nums: A list of integers
* @return: A list of integers includes the index of the first number
* and the index of the last number
*/
public int[] subarraySumClosest(int[] nums) {
int[] res = new int[2];
if (nums == null || nums.length == 0) {
return res;
}
int len = nums.length;
if(len == 1) {
res[0] = res[1] = 0;
return res;
}
Pair[] sums = new Pair[len+1];
int prev = 0;
sums[0] = new Pair(0, 0);
for (int i = 1; i <= len; i++) {
sums[i] = new Pair(prev + nums[i-1], i);
prev = sums[i].sum;
}
Arrays.sort(sums, new Comparator<Pair>() {
public int compare(Pair a, Pair b) {
return a.sum - b.sum;
}
});
int ans = Integer.MAX_VALUE;
for (int i = 1; i <= len; i++) {
if (ans > sums[i].sum - sums[i-1].sum) {
ans = sums[i].sum - sums[i-1].sum;
int[] temp = new int[]{sums[i].index - 1, sums[i - 1].index - 1};
Arrays.sort(temp);
res[0] = temp[0] + 1;
res[1] = temp[1];
}
}
return res;
}
}
2017-02-14
165.merge two sorted lists
要求：如题
思路：（自写）dummy node，和sorted array类似
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode lastNode = dummy;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
lastNode.next = l1;
l1 = l1.next;
} else {
lastNode.next = l2;
l2 = l2.next;
}
lastNode = lastNode.next;
}
if (l1 != null) {
lastNode.next = l1;
}
if (l2 != null) {
lastNode.next = l2;
}
return dummy.next;
}
2017-03-06
599.insert into a cyclic sorted list
要求：如题
思路：（自写但逻辑有错误）根据前后关系找插入的位置
public ListNode insert(ListNode node, int x) {
ListNode newNode = new ListNode(x);
if (node == null) {
newNode.next = newNode;
return newNode;
}
if (node.next == node) {
node.next = newNode;
newNode.next = node;
return newNode;
}
ListNode currNode = node;
ListNode nextNode = node.next;
while (currNode.val <= nextNode.val) {
if (nextNode.next.val <= nextNode.val) {
if (nextNode.val >= x && currNode.val <= x) {
currNode.next = newNode;
newNode.next = nextNode;
break;
} else {
newNode.next = nextNode.next;
nextNode.next = newNode;
break;
}
} else {
currNode = nextNode;
nextNode = nextNode.next;
}
}
return newNode;
}
答案：
public ListNode insert(ListNode node, int x) {
if (node == null) {
node = new ListNode(x);
node.next = node;
return node;
}
//找断点
ListNode curr = node;
ListNode prev = null;
do {
prev = curr;
curr = curr.next;
if (x <= curr.val && x >= prev.val) {
break;
}
if ((prev.val > curr.val) && (x < curr.val || x > prev.val)) {
break;
}
} while (curr != node);
//找到插入位置后连接
ListNode newNode = new ListNode(x);
newNode.next = curr;
prev.next = newNode;
return newNode;
}
2017-03-06
102.listed list cycle
要求：判断链表是否有环
思路：（修正答案错误）快慢指针，有环的话会相遇
warn：注意判决条件
public boolean hasCycle(ListNode head) {
if (head == null || head.next == null) {
return false;
}
//快慢指针
ListNode slow = head;
ListNode fast = head.next;
while (slow != fast) {
if (slow == null || fast == null || fast.next == null) {
return false;
}
slow = slow.next;
fast = fast.next.next;
}
return true;
}
2017-03-06
98.sort list
要求：把链表排序
思路：（答案）分别用归并和快速排序来完成，背
快速：
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;
}
归并：
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);
}
2017-03-07
105.copy list with random pointer
要求：深度复制带有随机指针的链表
思路：（看视频后自写）分三步：1、用next来保存新老节点的对应关系，形成1->1'->2->2'->3->3'的结构 2、复制random指针 3、拆分并各自连回去
warn：注意边界条件！
public RandomListNode copyRandomList(RandomListNode head) {
if (head == null) {
return null;
}
//three steps
copyNext(head);
copyRandom(head);
return spiltList(head);
}
private void copyNext(RandomListNode head) {
while (head != null) {
RandomListNode newNode = new RandomListNode(head.label);
newNode.next = head.next;
head.next = newNode;
head = head.next.next;
}
}
private void copyRandom(RandomListNode head) {
while (head != null) {
if (head.random != null) {
head.next.random = head.random.next;
}
head = head.next.next;
}
}
private RandomListNode spiltList(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;
}
2017-03-08
5.kth largest element(for 65)
要求：如题
思路：（小视频答案）quick select算法，思路和快速排序类似，找标杆把数排成两边
warn：注意每一次递归时候的取值
public int kthLargestElement(int k, int[] nums) {
if (nums == null || nums.length == 0) {
return -1;
}
return quickSelect(nums, 0, nums.length - 1, k);
}
private int quickSelect(int[] nums, int start, int end, int k) {
if (start == end) {
return nums[start];
}
int i = start, j = end;
int pivot = nums[(i + j) / 2];
while (i <= j) {
while (i <= j && nums[i] > pivot) {
i++;
}
while (i <= j && nums[j] < pivot) {
j--;
}
if (i <= j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
i++;
j--;
}
}
//三种情况，左中右，i和j中间有可能还有一个数刚好满足
if (start + k - 1 <= j) {
return quickSelect(nums, start, j, k);
}
if (start + k - 1 >= i) {
//右边的时候需要去掉左边，所以不是找第k大了
return quickSelect(nums, i, end, k - (i - start));
}
return nums[j + 1];
}
2017-03-09
65.median of two sorted array(hard)
要求：找出两个排序数组的中位数
思路：（看视频后自写）转换成求第k大的数，然后把k = len / 2代入（偶数时需要求len / 2和len / 2 + 1并取平均）时间复杂度推算法，在O(1)时间内，把求第k大的问题简化成第k / 2的问题，也就是对比两个数组k / 2位置上的数，小的那个就把该数组前面的数都丢掉（加入indexA和indexB来实现）
tips：若一个数组不够长取到index + k / 2 - 1个数，则用无穷大代替
warn：当index大于length时，证明第k大个数只存在于另一数组
public double findMedianSortedArrays(int[] A, int[] B) {
int len = A.length + B.length;
if (len % 2 == 0) {
return (findKth(A, 0, B, 0, len / 2) + findKth(A, 0, B, 0, len / 2 + 1)) / 2.0;
}
return findKth(A, 0, B, 0, len / 2 + 1);
}
private double findKth(int[] A, int indexA, int[] B, int indexB, int k) {
if (indexA >= A.length) {
return B[indexB + k - 1];
}
if (indexB >= B.length) {
return A[indexA + k - 1];
}
if (k == 1) {
return Math.min(A[indexA], B[indexB]);
}
//若数组不够长，则赋为无限大
int keyA = Integer.MAX_VALUE;
int keyB = Integer.MAX_VALUE;
if (indexA + k / 2 - 1 < A.length) {
keyA = A[indexA + k / 2 - 1];
}
if (indexB + k / 2 - 1 < B.length) {
keyB = B[indexB + k / 2 - 1];
}
if (keyA < keyB) {
return findKth(A, indexA + k / 2, B, indexB, k - k / 2);
} else {
return findKth(A, indexA, B, indexB + k / 2, k - k / 2);
}
}
2017-03-09
451.Swap Nodes in Pairs
要求：把链表两两交换
思路：建立哨兵节点dummyNode，每次指针停留在要交换节点的上一节点
warn：注意null判断
2017-09-12
35.Reverse Linked List 翻转链表
要求：如题
思路：创建前向指针（与dummy其实一样）prev，不用交换数字，而是改变链的方向，然后把两个指针逐渐向后移动
2017-09-12
450.Reverse Nodes in k-Group（hard）
要求：链表中每k个翻转一次，最后不满k个的不要翻转
思路：写一个函数，完成以下几点：1. 记录了每次开始翻转的上一个节点（一开始的时候便是dummy node），这样才可以实现拼接 2.数k次，看是否到结尾 3.翻转
翻转过程中需要记录的节点：1. 开始翻转的第一个节点（翻转后变最后一个点，用于拼接后面） 2. 翻转过程的前后指针prev和curt，临时记录curt.next的temp，因为翻转就是令curt.next = prev，也就是指向前一个节点，然后让prev = curt，curt = temp即把指针后移
public class Solution {
/*
* @param head: a ListNode
* @param k: An integer
* @return: a ListNode
*/
public ListNode reverseKGroup(ListNode head, int k) {
// write your code here
if (head == null || k <= 1) {
return head;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
head = dummy;
while (head.next != null) {
//记录每一次开始节点的上一个位置
head = reverseNextK(head, k);
}
return dummy.next;
}
private ListNode reverseNextK(ListNode head, int k) {
ListNode nk = head;
for (int i = 0; i < k; i++) {
//数k个，若未数完就出现null，则链表不需要再翻转
if (nk.next == null) {
return nk;
}
nk = nk.next;
}
//翻转过程
//1. head记录的是此次要翻转段的上一个节点,first记录第一个翻转的点
// prev,curt为翻转时的前后指针
ListNode first = head.next;
ListNode prev = null, curt = first;
for (int i = 0; i < k; i++) {
//保存
ListNode temp = curt.next;
//反向
curt.next = prev;
//指针移动
prev = curt;
curt = temp;
}
//拼接,第一个翻转的点接下一次需要开始翻转的点，原head接这次翻转最后的点
first.next = curt;
head.next = prev;
//first是下一次的head
return first;
}
}
2017-09-18
96.链表划分
要求：给定一个单链表和数值x，划分链表使得所有小于x的节点排在大于等于x的节点之前。
思路：重新做两个链表 用两个dummy 分别存小于的部分和大于的部分。
public class Solution {
/*
* @param head: The first node of linked list
* @param x: An integer
* @return: A ListNode
*/
public ListNode partition(ListNode head, int x) {
// write your code here
if (head == null) {
return null;
}
ListNode leftDummy = new ListNode(0);
ListNode rightDummy = new ListNode(0);
ListNode left = leftDummy;
ListNode right = rightDummy;
while (head != null) {
if (head.val < x) {
left.next = head;
left = left.next;
} else {
right.next = head;
right = right.next;
}
head = head.next;
}
//记得把最右边的与原先的断开
right.next = null;
left.next = rightDummy.next;
return leftDummy.next;
}
}
2017-10-15
547.两数组的交
要求：交中没有重复数
思路：用两个HashSet，第一个Set1把其中一个数组全部放入，Set2放入另一数组时判断自己没有且Set1有才放入，就可以求它们的交，且交中没有重复
public int[] intersection(int[] nums1, int[] nums2) {
// write your code here
if (nums1 == null || nums1.length == 0
|| nums2 == null || nums2.length == 0) {
return new int[0];
}
Set<Integer> set1 = new HashSet<>();
for (int num : nums1) {
if (!set1.contains(num)) {
set1.add(num);
}
}
Set<Integer> set2 = new HashSet<>();
for (int num : nums2) {
if (set1.contains(num) && !set2.contains(num)) {
set2.add(num);
}
}
int[] result = new int[set2.size()];
int i = 0;
for (int num : set2) {
result[i++] = num;
}
return result;
}
2018-01-17
548.两数组的交II
要求：交中有重复数，即两数组中若有相同的重复数也放进交中
思路：用HashMap记录其中第一个数组的数和出现次数，用List放入另一数组的数，并在放入前判断Map中是否有该数，且出现次数大于0，放入后把Map中的次数减1
public int[] intersection(int[] nums1, int[] nums2) {
// write your code here
if (nums1 == null || nums1.length == 0
|| nums2 == null || nums2.length == 0) {
return new int[0];
}
Map<Integer, Integer> map = new HashMap<>();
for (int num : nums1) {
if (map.containsKey(num)) {
map.put(num, map.get(num) + 1);
} else {
map.put(num, 1);
}
}
List<Integer> list = new ArrayList<>();
for (int num : nums2) {
if (map.containsKey(num) && map.get(num) > 0) {
list.add(num);
map.put(num, map.get(num) - 1);
}
}
int[] result = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
result[i] = list.get(i);
}
return result;
}
2018-01-17
6.合并排序数组
要求：合并排序数组变成一个新数组
思路：两个数组都还有数时就判断谁小谁先进，其中一个数组越界后就一直放另一个数组
public int[] mergeSortedArray(int[] A, int[] B) {
int m = A.length;
int n = B.length;
int[] C = new int[m + n];
int i = 0, j = 0, k = 0;
while (i < m && j < n) {
if (A[i] < B[j]) {
C[k++] = A[i++];
} else {
C[k++] = B[j++];
}
}
while (i < m) {
C[k++] = A[i++];
}
while (j < n) {
C[k++] = B[j++];
}
return C;
}
2018-01-17
64.合并排序数组II
要求：在A数组上扩展，其中A后面有预留位置
思路：从后面开始保存就不会覆盖到A前面的数
public void mergeSortedArray(int[] A, int m, int[] B, int n) {
// write your code here
int i = m - 1, j = n - 1, index = m + n - 1;
while (i >= 0 && j >= 0) {
if (A[i] >= B[j]) {
A[index--] = A[i--];
} else {
A[index--] = B[j--];
}
}
while (i >= 0) {
A[index--] = A[i--];
}
while (j >= 0) {
A[index--] = B[j--];
}
}
2018-01-17
7 - Two Pointers
607.two sum - data structure
要求：创造一个数据结构，有添加新元素、查找TwoSum的功能
思路：（自写）开两个全局变量list和map，用map的原因是防止有两个重复数加起来满足TwoSum的target，这时候可以把map的value记录key的个数
warn：记得写构造方法！！！
public class TwoSum {
private ArrayList<Integer> list;
private Map<Integer, Integer> map;
public TwoSum() {
list = new ArrayList<Integer>();
map = new HashMap<Integer, Integer>();
}
public void add(int number) {
if (map.containsKey(number)) {
map.put(number, map.get(number) + 1);
} else {
map.put(number, 1);
list.add(number);
}
}
// Find if there exists any pair of numbers which sum is equal to the value.
public boolean find(int value) {
if (list == null || list.size() == 0) {
return false;
}
for (int i = 0; i < list.size(); i++) {
int num1 = list.get(i), num2 = value - list.get(i);
if ((num1 == num2 && map.get(num1) > 1) || (num1 != num2 && map.containsKey(num2))) {
return true;
}
}
return false;
}
}
// Your TwoSum object will be instantiated and called as such:
// TwoSum twoSum = new TwoSum();
// twoSum.add(number);
// twoSum.find(value);
2017-03-12
521.remove duplicate numbers in array(去重)
要求：如题
思路1：（自写AC）判断与前一数是否一样，不一样再动index
warn：记得排序才能这样做！！
public int deduplication(int[] nums) {
if (nums == null || nums.length == 0) {
return 0;
}
Arrays.sort(nums);
int index = 1;
for (int i = 1; i < nums.length; i++) {
if (nums[i] != nums[i - 1]) {
nums[index++] = nums[i];
}
}
return index;
}
思路2：hashmap判断是否存在（答案）
public int deduplication(int[] nums) {
HashMap<Integer, Boolean> mp = new HashMap<Integer, Boolean>();
for (int i = 0; i < nums.length; ++i)
mp.put(nums[i], true);
int result = 0;
for (Map.Entry<Integer, Boolean> entry : mp.entrySet())
nums[result++] = entry.getKey();
return result;
}
2017-03-09
609.two sum - less than or equal to target
要求：如题
思路1：（自写没AC！）同向双指针，即使经过优化后，数据大了还会TLE
public int twoSum5(int[] nums, int target) {
if (nums == null || nums.length < 2) {
return 0;
}
Arrays.sort(nums);
int number = 0;
int end = nums.length;
for (int i = 0; i < nums.length - 1; i++) {
if (i >= end) {
break;
}
for (int j = i + 1; j < end; j++) {
if (nums[i] + nums[j] <= target) {
number++;
} else {
end = j;
break;
}
}
}
return number;
}
思路2：（答案思路）相向指针，一次过算完中间有几个满足条件的数
public int twoSum5(int[] nums, int target) {
if (nums == null || nums.length < 2) {
return 0;
}
Arrays.sort(nums);
int number = 0;
int i = 0, j = nums.length - 1;
while (i < j) {
if (nums[i] + nums[j] > target) {
j--;
} else {
number += j - i;
i++;
}
}
return number;
}
2017-03-10
608.two sum - input array is sorted
要求：如题
思路1：（自写AC）HashMap判断是否存在target - nums[i]，但这样没用到排序特性
public int[] twoSum(int[] nums, int target) {
if (nums == null || nums.length < 2) {
return new int[2];
}
Map<Integer, Integer> map = new HashMap<>();
int[] res = new int[2];
for (int i = 0; i < nums.length; i++) {
if (map.containsKey(target - nums[i])) {
res[0] = map.get(target - nums[i]) + 1;
res[1] = i + 1;
break;
}
map.put(nums[i], i);
}
return res;
}
思路2：（自写AC）two pointers，相向
public int[] twoSum(int[] nums, int target) {
if (nums == null || nums.length < 2) {
return new int[2];
}
int[] res = new int[2];
int i = 0, j = nums.length - 1;
while (i < j) {
int sum = nums[i] + nums[j];
if (sum < target) {
i++;
} else if (sum > target) {
j--;
} else {
res[0] = i + 1;
res[1] = j + 1;
break;
}
}
return res;
}
2017-03-10
587.two sum - unique pairs
要求：求所有和为target的数量
思路：（自写差点一次AC）还是相向指针，没AC的原因是if(target == sum)时忘记判断i < j，导致多了一个
public int twoSum6(int[] nums, int target) {
if (nums == null || nums.length < 2) {
return 0;
}
Arrays.sort(nums);
int number = 0;
int i = 0, j = nums.length - 1;
while (i < j) {
while (j != nums.length - 1 && i < j && nums[j] == nums[j + 1]) {
j--;
}
while (i != 0 && i < j && nums[i] == nums[i - 1]) {
i++;
}
int sum = nums[i] + nums[j];
if (i < j && sum == target) {
number++;
i++;
j--;
} else if (sum < target) {
i++;
} else {
j--;
}
}
return number;
}
2017-03-10
533.two sum - closest to target
要求：求最接近的差
思路：（自写AC）加入diff变量，仍是套用相向指针模板
public int twoSumClosest(int[] nums, int target) {
if (nums == null || nums.length < 2) {
return -1;
}
Arrays.sort(nums);
int i = 0, j = nums.length - 1;
int diff = Integer.MAX_VALUE;
while (i < j) {
int sum = nums[i] + nums[j];
diff = Math.min(diff, Math.abs(sum - target));
if (target == sum) {
return 0;
} else if (target < sum) {
j--;
} else {
i++;
}
}
return diff;
}
2017-03-10
148.sort colors
要求：把0，1，2排序
思路：（答案）三个指针！！添加一个中间指针i。左边交换时i和left同时动，右边交换时只有right动，不交换i自己动
warn：判断条件是i <= right
public void sortColors(int[] nums) {
if (nums == null || nums.length < 2) {
return;
}
int i = 0, left = 0, right = nums.length - 1;
while (i <= right) {
if (nums[i] == 0) {
swap(nums, left, i);
i++;
left++;
} else if (nums[i] == 1) {
i++;
} else {
swap(nums, right, i);
right--;
}
}
}
private void swap(int[] nums, int a, int b) {
int temp = nums[a];
nums[a] = nums[b];
nums[b] = temp;
}
2017-03-10
143.sort colors II
要求：有k种颜色，对n个对象进行排序
思路：（答案）先推出时间复杂度为nlogk，然后对k种颜色使用快排思想，称为rainbowSort彩虹排序
public void sortColors2(int[] colors, int k) {
if (colors == null || colors.length == 0) {
return;
}
rainbowSort(colors, 0, colors.length - 1, 1, k);
}
public void rainbowSort(int[] colors,
int left,
int right,
int colorFrom,
int colorTo) {
if (colorFrom == colorTo) {
return;
}
if (left >= right) {
return;
}
int colorMid = (colorFrom + colorTo) / 2;
int l = left, r = right;
while (l <= r) {
while (l <= r && colors[l] <= colorMid) {
l++;
}
while (l <= r && colors[r] > colorMid) {
r--;
}
if (l <= r) {
int temp = colors[l];
colors[l] = colors[r];
colors[r] = temp;
l++;
r--;
}
}
rainbowSort(colors, left, r, colorFrom, colorMid);
rainbowSort(colors, l, right, colorMid + 1, colorTo);
}
2017-03-12
57.3Sum
要求：存在重复数，相加为0
思路1：（自写AC）化为two sum，用set去重答案
public ArrayList<ArrayList<Integer>> threeSum(int[] numbers) {
ArrayList<ArrayList<Integer>> results = new ArrayList<>();
Set<ArrayList<Integer>> set = new HashSet<>();
Arrays.sort(numbers);
for (int i = 0; i < numbers.length - 2; i++) {
int left = i + 1, right = numbers.length - 1;
while (left < right) {
int sum = numbers[left] + numbers[right];
if (sum == -numbers[i]) {
ArrayList<Integer> list = new ArrayList<>();
list.add(numbers[i]);
list.add(numbers[left]);
list.add(numbers[right]);
if (!set.contains(list)) {
results.add(list);
set.add(list);
}
left++;
right--;
} else if (sum < -numbers[i]) {
left++;
} else {
right--;
}
}
}
return results;
}
思路2：（答案）排序后用number[i] == number[i - 1]去重，更省空间
public ArrayList<ArrayList<Integer>> threeSum(int[] nums) {
ArrayList<ArrayList<Integer>> results = new ArrayList<>();
if (nums == null || nums.length < 3) {
return results;
}
Arrays.sort(nums);
for (int i = 0; i < nums.length - 2; i++) {
// skip duplicate triples with the same first numebr
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int left = i + 1, right = nums.length - 1;
int target = -nums[i];
twoSum(nums, left, right, target, results);
}
return results;
}
public void twoSum(int[] nums,
int left,
int right,
int target,
ArrayList<ArrayList<Integer>> results) {
while (left < right) {
if (nums[left] + nums[right] == target) {
ArrayList<Integer> triple = new ArrayList<>();
triple.add(-target);
triple.add(nums[left]);
triple.add(nums[right]);
results.add(triple);
left++;
right--;
// skip duplicate pairs with the same left
while (left < right && nums[left] == nums[left - 1]) {
left++;
}
// skip duplicate pairs with the same right
while (left < right && nums[right] == nums[right + 1]) {
right--;
}
} else if (nums[left] + nums[right] < target) {
left++;
} else {
right--;
}
}
}
2017-03-12
31.partition array
要求：以k为标准把数组分成两边
思路：（自写但有bug）quick select思想，相向指针把小于的放一边大于放另一边
warn：注意边界条件为left <= right才对！！
public int partitionArray(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return 0;
}
int left = 0, right = nums.length - 1;
while (left <= right) {
while (left <= right && nums[left] < k) {
left++;
}
while (left <= right && nums[right] >= k) {
right--;
}
if (left <= right) {
int temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
left++;
right--;
}
}
return left;
}
2017-03-12
604.滑动窗口内数的和
要求：给定大小为n的数组和大小为k的滑动窗口，输出从数组开头滑动每一个时刻的窗口内数的和
思路：先求出开始第一个窗口的和sum，然后每移动一次减去头加上尾的数
public int[] winSum(int[] nums, int k) {
if (nums == null || k == 0 || nums.length < k) {
return new int[0];
}
int[] sums = new int[nums.length - k + 1];
int sum = 0;
for (int index = 0; index < k; index++) {
sum += nums[index];
}
sums[0] = sum;
int i = 0;
int j = k - 1;
while (j < nums.length - 1) {
sum = sum - nums[i++] + nums[++j];
sums[i] = sum;
}
return sums;
}
2018-01-17
539.移动零
要求：把数组中的0都移到尾部，其他数保持原有顺序
思路：首先把不是0的数按顺序赋值到数组前面，然后再在数组剩下的部分补充0
public void moveZeroes(int[] nums) {
// write your code here
if (nums == null || nums.length == 0) {
return;
}
int index = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] != 0) {
nums[index++] = nums[i];
}
}
while (index < nums.length) {
nums[index++] = 0;
}
}
2018-01-17
415.有效回文串
要求：只包含数字和字母，忽略大小写
思路：首尾指针开始先判断是不是数字或者字母，若是则统一转成小写（或者大写）来比较
public boolean isPalindrome(String s) {
// write your code here
if (s == null || s.length() == 0) {
return true;
}
int start = 0;
int end = s.length() - 1;
while (start < end) {
while (start < s.length() && !isValid(s.charAt(start))) {
start++;
}
while (end >= 0 && !isValid(s.charAt(end))) {
end--;
}
if (start < s.length() && end >= 0 && Character.toLowerCase(s.charAt(start))
!= Character.toLowerCase(s.charAt(end))) {
return false;
}
start++;
end--;
}
return true;
}
private boolean isValid(char c) {
return Character.isLetter(c) || Character.isDigit(c);
}
2018-01-17
56.两数之和
要求：在数组中找出两数的和等于指定数target，返回数组对应的下标
思路：用HashMap的key装数组的值，用于判断是否存在两数之和等于target，用value装下标
public int[] twoSum(int[] numbers, int target) {
// write your code here
if (numbers == null || numbers.length == 0) {
return new int[]{0, 0};
}
Map<Integer, Integer> map = new HashMap<>();
int[] result = new int[2];
for (int i = 0; i < numbers.length; i++) {
if (map.containsKey(target - numbers[i])) {
result[0] = map.get(target - numbers[i]);
result[1] = i + 1;
break;
}
map.put(numbers[i], i + 1);
}
return result;
}
2018-01-17
8 - 哈希表与堆
128.hash function
要求：按照题目给出的公式算
思路：（自写一直不成功，不知原因）最重要是字符转ascii码
public int hashCode(char[] key, int HASH_SIZE) {
if (key == null || key.length == 0) {
return 0;
}
long ans = 0;
for(int i = 0; i < key.length;i++) {
ans = (ans * 33 + (int)(key[i])) % HASH_SIZE;
}
return (int)ans;
}
2017-03-13
129.rehashing
要求：把哈希表复制一遍并增大一倍size
思路：（自写时没有考虑好ListNode[]中存的到底是实体还是指针，其实是指针）deep copy时记住若原本的hashTable[i] !=null，它仍可能有下一项，而且复制过去的newTable[i]，它不一定是第一次用i这个位置，若不是的时候，首先要用dummy指针获得链表最后的位置
Notice：（题目自带）if you directly calculate -4 % 3 you will get -1. You can use function: a % b = (a % b + b) % b to make it is a non negative integer
public ListNode[] rehashing(ListNode[] hashTable) {
// write your code here
if (hashTable.length <= 0) {
return hashTable;
}
int newcapacity = 2 * hashTable.length;
ListNode[] newTable = new ListNode[newcapacity];
for (int i = 0; i < hashTable.length; i++) {
while (hashTable[i] != null) {
int newindex
= (hashTable[i].val % newcapacity + newcapacity) % newcapacity;
if (newTable[newindex] == null) {
newTable[newindex] = new ListNode(hashTable[i].val);
// newTable[newindex].next = null;
} else {
ListNode dummy = newTable[newindex];
while (dummy.next != null) {
dummy = dummy.next;
}
dummy.next = new ListNode(hashTable[i].val);
}
hashTable[i] = hashTable[i].next;
}
}
return newTable;
}
2017-03-16
134.LRU cache(hard)
要求：实现一个数据结构，使满足1. get方法，把key对应的value取出，并把key放大最后（代表最近使用，也就是LRU） 2. set方法，若存在key则更新value，并把key放到最后，若超过此数据结构的capacity，要把头部的数据删掉，再在末尾插入
思路：（答案+自己注释）要点 1. 定义双向链表node 2.LRU方案需要head、tail节点支持后面的操作，并用hashmap来链接key和node的位置(这样可以通过O1的时间找到node)，自己包含一个最大长度capacity 3.get方法中记得判断有无，有的情况下记得把原来key的位置断开并连接前后，然后把key放到结尾 4.set方法中判断存不存在key的时候要用get方法判断（不能直接用map判断，因为用get方法在存在key时就可以顺便完成了断开中间并移到末尾的操作），不存在的时候则考虑有没有超过capacity，超过就先去掉头部再在末尾插入
详细看注释：
public class Solution {
//双向链表
private class Node {
Node prev;
Node next;
int key;
int value;
//内部类的构造方法
public Node(int key, int value) {
this.key = key;
this.value = value;
this.prev = null;
this.next = null;
}
}
//在LRU方案中 需要头尾节点来辅助更改 不存储真正的数据
//key与node在map中匹配 这样可以通过key找到node
private int capacity;
private Map<Integer, Node> map = new HashMap<>();
private Node head = new Node(-1, -1);
private Node tail = new Node(-1, -1);
// @param capacity, an integer
//LRU cache的构造函数
public Solution(int capacity) {
this.capacity = capacity;
head.next = tail;
tail.prev = head;
}
// @return an integer
//1. 判断有无
//2. 有的话取出它的value 并移到末尾(表示最近使用过)
public int get(int key) {
if (!map.containsKey(key)) {
return -1;
}
//用hashmap定位node在哪 然后移除它
Node current = map.get(key);
current.prev.next = current.next;
current.next.prev = current.prev;
//把它移到末尾
moveToTail(current);
return map.get(key).value;
}
// @param key, an integer
// @param value, an integer
// @return nothing
//1. 若已存在key 则更新它的value
//2. 若到达LRU cache的最大capacity 将头部删除并插入在最后
public void set(int key, int value) {
//存在时更新value 并通过get方法讲key移到最后！！！
if (get(key) != -1) {
map.get(key).value = value;
return;
}
if (map.size() == capacity) {
map.remove(head.next.key);
head.next = head.next.next;
head.next.prev = head;
}
Node newNode = new Node(key, value);
map.put(key, newNode);
moveToTail(newNode);
}
private void moveToTail(Node node) {
node.prev = tail.prev;
tail.prev = node;
node.prev.next = node;
node.next = tail;
}
}
2017-03-16
544.top k largest numbers
要求：把前k大的数输出
思路：（自写，注意写法）PriorityQueue最基本用法，java中默认的是实现了minheap的优先队列，而求最大的数用的是最大堆，注意构造PQ时候的comparator写法，这样可以实现最大堆，而队列长就是k
public int[] topk(int[] nums, int k) {
PriorityQueue<Integer> minheap = new PriorityQueue<Integer>(k, new Comparator<Integer>() {
public int compare(Integer o1, Integer o2) {
if (o1 < o2) {
return 1;
} else if (o1 > o2) {
return -1;
} else {
return 0;
}
}
});
for (int i : nums) {
minheap.add(i);
}
int[] res = new int[k];
for (int i = 0; i < res.length; i++) {
res[i] = minheap.poll();
}
return res;
}
2017-03-16
606.kth largest element II
要求：把第k大数输出就行
思路1：（自写AC）同上，这次只需要输出一个数
public int kthLargestElement2(int[] nums, int k) {
if (nums == null || nums.length == 0 || k == 0) {
return 0;
}
PriorityQueue<Integer> minheap = new PriorityQueue<>(k, new Comparator<Integer>() {
public int compare(Integer o1, Integer o2) {
if (o1 < o2) {
return 1;
} else if (o1 > o2) {
return -1;
} else {
return 0;
}
}
});
for (int i : nums) {
minheap.add(i);
}
int res = 0;
for (int i = 0; i < k; i++) {
if (i == k - 1) {
res = minheap.peek();
}
minheap.poll();
}
return res;
}
思路2：也可以用quickselect来做
2017-03-17
612.k closest point
要求：求到origin坐标最近的k个点
思路：（答案）还是PQ的应用，不过仍不知道PQ里面的comparator到底怎么设计，答案中设计的是以diff为参考值的最大堆！！
warn：不知道怎么搞的，明明限制了pq的长度是k，但超过k时候仍然不会自动丢弃，而要有pq.size() > k时poll()，而且不把origin点设成全局变量会报错...
public class Solution {
/**
* @param points a list of points
* @param origin a point
* @param k an integer
* @return the k closest points
*/
private Point global_origin = null;
public Point[] kClosest(Point[] points, Point origin, int k) {
global_origin = origin;
PriorityQueue<Point> pq = new PriorityQueue<>(k, new Comparator<Point>() {
public int compare(Point a, Point b) {
int diff = getDistance(b, global_origin) - getDistance(a, global_origin);
if (diff == 0) {
diff = b.x - a.x;
}
if (diff == 0) {
diff = b.y - a.y;
}
return diff;
}
});
for (Point p : points) {
pq.add(p);
//最大堆 超过长度的poll掉
if (pq.size() > k) {
pq.poll();
}
}
Point[] res = new Point[k];
while (!pq.isEmpty()) {
res[--k] = pq.poll();
}
return res;
}
private int getDistance(Point a, Point b) {
return (a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y);
}
}
2017-03-17
613.high five
要求：找出学生id对应的前五位成绩，求平均后放到map中与id对应起来
思路：（自写答案不对？）用之前的思路，pq判断前五大成绩，然后求平均
public class Solution {
/**
* @param results a list of <student_id, score>
* @return find the average of 5 highest scores for each person
* Map<Integer, Double> (student_id, average_score)
*/
public Map<Integer, Double> highFive(Record[] results) {
if (results == null || results.length == 0) {
return new HashMap<Integer, Double>();
}
Map<Integer, PriorityQueue<Integer>> map = new HashMap<>();
for (int i = 0; i < results.length; i++) {
if (!map.containsKey(results[i].id)) {
map.put(results[i].id, new PriorityQueue<Integer>(
5, new Comparator<Integer>() {
public int compare(Integer o1, Integer o2) {
if (o1 < o2) {
return 1;
} else if (o1 > o2) {
return -1;
} else {
return 0;
}
}
}));
} else {
map.get(results[i].id).add(results[i].score);
}
}
Map<Integer, Double> res = new HashMap<>();
for (int i = 0; i < results.length; i++) {
if (!res.containsKey(results[i].id)) {
double average = getAverage(map.get(results[i].id));
res.put(results[i].id, average);
}
}
return res;
}
private double getAverage(PriorityQueue<Integer> pq) {
double sum = 0;
while (!pq.isEmpty()) {
double score = pq.poll();
sum += score;
}
return sum / 5.0;
}
}
答案：直接用ArrayList来存成绩，因为只取五项比较少，所以可以通过循环来获得最小的来和目前的比较，比最小的大就存进去（其实这也是堆的原理，不过如果取的项比较多，这种方法应该不行）
public class Solution {
/**
* @param results a list of <student_id, score>
* @return find the average of 5 highest scores for each person
* Map<Integer, Double> (student_id, average_score)
*/
public Map<Integer, Double> highFive(Record[] results) {
Map<Integer, Double> answer = new HashMap<Integer, Double>();
Map<Integer, List<Integer>> hash = new HashMap<Integer, List<Integer>>();
for (Record r : results) {
if (!hash.containsKey(r.id)){
hash.put(r.id, new ArrayList<Integer>());
}
if (hash.get(r.id).size() < 5) {
hash.get(r.id).add(r.score);
} else {
int index = 0;
for (int i = 1; i < 5; ++i)
if (hash.get(r.id).get(i) < hash.get(r.id).get(index))
index = i;
if (hash.get(r.id).get(index) < r.score)
hash.get(r.id).set(index, r.score);
}
}
for (Map.Entry<Integer, List<Integer>> entry : hash.entrySet()) {
int id = entry.getKey();
List<Integer> scores = entry.getValue();
double average = 0;
for (int i = 0; i < 5; ++i)
average += scores.get(i);
average /= 5.0;
answer.put(id, average);
}
return answer;
}
}
2017-03-17
104.merge k sorted lists
要求：合并k个有序链表
思路：（看答案自写）1. 把所有链表头放进heap 2. 把当前链表头的next放进去heap（这样又自动排序了）有点像BFS的思想
自写时把链表头放进heap时忘记判断lists.get(i) != null
public class Solution {
/**
* @param lists: a list of ListNode
* @return: The head of one sorted list.
*/
private Comparator<ListNode> ListNodeComparator = new Comparator<ListNode>() {
public int compare(ListNode left, ListNode right) {
return left.val - right.val;
}
};
public ListNode mergeKLists(List<ListNode> lists) {
if (lists == null || lists.size() == 0) {
return null;
}
Queue<ListNode> heap = new PriorityQueue<>(lists.size(), ListNodeComparator);
//把所有链表头先放入heap
for (int i = 0; i < lists.size(); i++) {
if (lists.get(i) != null) {
heap.add(lists.get(i));
}
}
ListNode dummy = new ListNode(0);
ListNode tail = dummy;
while (!heap.isEmpty()) {
ListNode current = heap.poll();
tail.next = current;
tail = current;
//若当前还有next 则添加到heap中继续排序
if (current.next != null) {
heap.add(current.next);
}
}
return dummy.next;
}
}
还可以用分治法来做
2017-03-17
4.丑数II
要求：找出只含素因子2,3,5的第n小的数，符合条件的数为1,2,3,4,5,6,8,9,10
思路：用三个指针来判断
public int nthUglyNumber(int n){
List<Integer> uglys = new ArrayList<Integer>(); //创建整形动态数组;
uglys.add(1); //把特殊值1首先加入;
int p2 = 0, p3 = 0, p5 = 0; //分别代表了与2、3、5相乘元素的序号,相当于指针;
//用前面各个位置的数与2,3,5相乘直到比上一个数大，并获得相应位置;
for (int i = 0; i <= n - 2; i++){
int lastNumber = uglys.get(i);
//获取数组中上一个元素，第n个数的上一个对应下标为n-2;
while (uglys.get(p2) * 2 <= lastNumber) p2++;
while (uglys.get(p3) * 3 <= lastNumber) p3++;
while (uglys.get(p5) * 5 <= lastNumber) p5++;
//对比p2,p3,p5位置上分别与2,3,5相乘的结果，最小值添加为下一元素;
uglys.add(Math.min(Math.min(uglys.get(p2) * 2, uglys.get(p3) * 3), uglys.get(p5) * 5));
}
return uglys.get(n - 1);
}
2018-01-17
594.strStr II
要求：用O(m + n)的时间复杂度完成查找字符串
思路：（小视频）Rabin Karp算法，利用hashcode来判断，若相同的时候才进行每一位的比较
warn：注意下标
public class Solution {
//hashcode
public final int BASE = 1000000;
/**
* @param source a source string
* @param target a target string
* @return an integer as index
*/
public int strStr2(String source, String target) {
if (source == null || target == null) {
return -1;
}
int m = target.length();
if (m == 0) {
return 0;
}
// count 31 ^ m
int power = 1;
for (int i = 0; i < m; i++) {
power = (power * 31) % BASE;
}
// target hashcode
int targetCode = 0;
for (int i = 0; i < m; i++) {
targetCode = (targetCode * 31 + target.charAt(i)) % BASE;
}
// substring hashcode
int substrCode = 0;
for (int i = 0; i < source.length(); i++) {
substrCode = (substrCode * 31 + source.charAt(i)) % BASE;
if (i < m - 1) {
continue;
}
// 超过m长度后减去头字符的hashcode
if (i >= m) {
substrCode = substrCode - (power * source.charAt(i - m)) % BASE;
if (substrCode < 0) {
substrCode += BASE;
}
}
if (substrCode == targetCode) {
if (source.substring(i - m + 1, i + 1).equals(target)) {
return i - m + 1;
}
}
}
return -1;
}
}
2017-03-22
134.LRU缓存策略（hard）
要求：为最近最小使用（LRU）缓存策略设计一个数据结构，支持get获取数据和set写入数据
get（key）：如果缓存存在key，则获取它的值，否则返回-1
set（key, value）：如果key没在缓存中则直接写入；当缓存达到上限，在写入数据前删除最近最少使用的数据来腾出空间
答案：
//双向链表
private class Node {
Node prev;
Node next;
int key;
int value;
//内部类的构造方法
public Node(int key, int value) {
this.key = key;
this.value = value;
this.prev = null;
this.next = null;
}
}
//在LRU方案中 需要头尾节点来辅助更改 不存储真正的数据
//key与node在map中匹配 这样可以通过key找到node
private int capacity;
private Map<Integer, Node> map = new HashMap<>();
private Node head = new Node(-1, -1);
private Node tail = new Node(-1, -1);
// @param capacity, an integer
//LRU cache的构造函数
public Solution(int capacity) {
this.capacity = capacity;
head.next = tail;
tail.prev = head;
} // @return an integer
//1. 判断有无
//2. 有的话取出它的value 并移到末尾(表示最近使用过)
public int get(int key) {
if (!map.containsKey(key)) {
return -1;
}
//用hashmap定位node在哪 然后移除它
Node current = map.get(key);
current.prev.next = current.next;
current.next.prev = current.prev;
//把它移到末尾
moveToTail(current);
return map.get(key).value;
} // @param key, an integer
// @param value, an integer
// @return nothing
//1. 若已存在key 则更新它的value
//2. 若到达LRU cache的最大capacity 将头部删除并插入在最后
public void set(int key, int value) {
//存在时更新value
if (get(key) != -1) {
map.get(key).value = value;
return;
}
if (map.size() == capacity) {
map.remove(head.next.key);
head.next = head.next.next;
head.next.prev = head;
}
Node newNode = new Node(key, value);
map.put(key, newNode);
moveToTail(newNode);
}
private void moveToTail(Node node) {
node.prev = tail.prev;
tail.prev = node;
node.prev.next = node;
node.next = tail;
}
2018-01-17
九章lintcode作业题的更多相关文章
- （lintcode全部题目解答之）九章算法之算法班题目全解(附容易犯的错误)
--------------------------------------------------------------- 本文使用方法:所有题目,只需要把标题输入lintcode就能找到.主要是 ...
- Python之路【第十九章】：Django进阶
Django路由规则 1.基于正则的URL 在templates目录下创建index.html.detail.html文件 <!DOCTYPE html> <html lang=&q ...
- 第十九章——使用资源调控器管理资源（1）——使用SQLServer Management Studio 配置资源调控器
原文:第十九章--使用资源调控器管理资源(1)--使用SQLServer Management Studio 配置资源调控器 本系列包含: 1. 使用SQLServer Management Stud ...
- 第十九章——使用资源调控器管理资源（2）——使用T-SQL配置资源调控器
原文:第十九章--使用资源调控器管理资源(2)--使用T-SQL配置资源调控器 前言: 在前一章已经演示了如何使用SSMS来配置资源调控器.但是作为DBA,总有需要写脚本的时候,因为它可以重用及扩展. ...
- 第十九章 Django的ORM映射机制
第十九章 Django的ORM映射机制 第一课 Django获取多个数据以及文件上传 1.获取多选的结果(checkbox,select/option)时: req.POST.getlist('fav ...
- Gradle 1.12用户指南翻译——第四十九章. Build Dashboard 插件
本文由CSDN博客貌似掉线翻译,其他章节的翻译请参见: http://blog.csdn.net/column/details/gradle-translation.html 翻译项目请关注Githu ...
- Gradle 1.12翻译——第十九章. Gradle 守护进程
有关其他已翻译的章节请关注Github上的项目:https://github.com/msdx/gradledoc/tree/1.12,或访问:http://gradledoc.qiniudn.com ...
- Gradle 1.12用户指南翻译——第二十九章. Checkstyle 插件
其他章节的翻译请参见: http://blog.csdn.net/column/details/gradle-translation.html 翻译项目请关注Github上的地址: https://g ...
- Gradle 1.12用户指南翻译——第三十九章. IDEA 插件
本文由CSDN博客万一博主翻译,其他章节的翻译请参见: http://blog.csdn.net/column/details/gradle-translation.html 翻译项目请关注Githu ...
随机推荐
- MVC源码分析 - Controller创建和创建扩展
上一篇, 出现了一个至关重要的类:MvcHandler, 接下来就来看一下MvcHandler吧. 先不看具体方法, 先看一下类里面的情况. //这里实现了两个重要的接口, 异步处理和同步处理的接口p ...
- RHEL账号总结一：账号的分类
账号是一种用来记录单个用户或者多个用户的数据.RHEL中每一个合法的用户都必须拥有账号,才能使用RHEL. 在RHEL上的账号可以分为两类: 用户账号:用来存储单一用户的数据,你也可以使用一个用户账号 ...
- 选择Blobs (Evision)
C++ // Sort by decreasing area Blobs.SortObjectsUsingFeature(OBJ_AREA, OBJ_SORT_DESCENDING); // Enum ...
- Maven配置插件跳过测试代码的编译和运行
Maven配置插件跳过测试代码的编译和运行: <!-- 编译插件 --> <plugin> <groupId>org.apache.maven.plugins< ...
- Codeforces Round #386 (Div. 2)
迟到的一次比赛 最近状态很崩溃 网速很慢 然后前面五题看了都有打 但是 只有A B E 是过了的 是时候要反省一下 A.随便判断一下就好 最少的份数嘛 B.画出来之后是一下子左边一下子右边 打一个递归 ...
- maven之（六）setting.xml的配置文件详解
setting.xml配置文件 maven的配置文件settings.xml存在于两个地方: 1.安装的地方:${M2_HOME}/conf/settings.xml 2.用户的目录:${user.h ...
- sharedMesh变量
在unity引擎中, 一般情况不建议用sharedMesh变量去写操作,建议只读,因为那会对mesh永久改变 变
- centos 6.5 安装openssl
1.下载wget https://www.openssl.org/source/openssl-1.0.2h.tar.gz 2.解压tar zxf openssl-1.0.2h.tar.gzcd op ...
- JavaScript进阶（一）
OK接下来,我们再次梳理一遍js并且提高一个等级. 众所周知,web前端开发者需要了解html和css,会只用html和css创建一个漂亮的页 面,但是这肯定是不够的,因为它只是一个静态的页面,我们 ...
- Unity2D开发小细节
当某个触碰物体挂在父组件时,如果当前子组件不加rigidbody,会默认使用父组件的rigidbody