从一道Hard学习滑动窗口
滑动窗口
滑动窗口(sliding windows algorithm)这种方法,专门用于解决区间解的问题。它在运算的时候,将解集放在窗口中,结束的时候比对是否符合预期。在运算的过程中,会对窗口的左右边缘进行操作(扩大、缩小)。特别针对于线性输入解决,滑动窗口就很形象了。
30. Substring with Concatenation of All Words
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.
题意是从一串输入字符串s中,找到能全包含words数组的起点(数组元素等长),这个起点可能有多个,而且不需要关心顺序。
Input:
s = "barfoothefoobarman",
words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoo" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
比如这个例子,他在第0个位置和第9个位置可以包含words数组,也就是0位置和9位置各有一个长度为6的窗口可以囊括words数组。
Template = "foo","bar" i= [b a r] f o o t h e f o o b a r m a n [b a r f o o] t h e f o o b a r m a n SAVE b a r f o o[] t h e f o o b a r m a n b a r f o o [t h e] f o o b a r m a n b a r f o o t h e [f o o] b a r m a n b a r f o o t h e [f o o b a r] m a n SAVE b a r f o o t h e f o o b a r[] m a n b a r f o o t h e f o o b a r [m a n] i= b [a r f] o o t h e f o o b a r m a n
.... i= b a [r f o] o t h e f o o b a r m a n
....
每次扩展使用单词列表中的词长,如果扩展过程中不符合预期则清除窗口,窗口左端在当前词总数大于词组总数的时候,计算总数是单词的长度(因为s除以单词长度余数是0到单词长度之间,覆盖了余数就能覆盖整个场景)。
public class FindSubstring {
public void test(){
// [0,9]
String s1 = "barfoothefoobarman";
String[] words1 = {"foo","bar"};
System.out.println(findSubstring(s1,words1));
// []
String s2 = "wordgoodgoodgoodbestword";
String[] words2 = {"word","good","best","word"};
System.out.println(findSubstring(s2,words2));
}
/**
* 单词等长
*/
public List<Integer> findSubstring(String s, String[] words) {
List<Integer> result = new ArrayList<>();
if(s == null || words.length == 0){
return result;
}
int step = words[0].length();
Map<String,Integer> counter = new HashMap<>();
for(String word :words){
counter.merge(word, 1, (a, b) -> a + b);
} for(int i=0;i<step;++i){
Map<String,Integer> window = new HashMap<>();
int left = i;
int right = i;
while (right <= s.length() - step && left <= s.length() - step*words.length){
String sub = s.substring(right,right + step);
window.merge(sub,1,(a , b)->a + b);
if(!counter.containsKey(sub)){
window.clear();
right += step;
left = right;
continue;
}
while (window.get(sub) > counter.get(sub)){
String drop = s.substring(left,left+step);
Integer dec = window.get(drop);
if(dec != null){
if(dec<=1){
window.remove(drop);
}else {
window.put(drop,dec-1);
}
}
left += step;
}
right += step;
if(right - left == step * words.length){
result.add(left);
}
}
}
return result;
} }
76. Minimum Window Substring
Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
Example:
Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC" 题意是将s子串中包含T的所有字母的最短子串找出来,例子中BANC是包含ABC的最短子串。
Template = "ABC" []A D O B E C O D E B A N C
[A] D O B E C O D E B A N C
[A D] O B E C O D E B A N C
[A D O] B E C O D E B A N C
[A D O B] E C O D E B A N C
[A D O B E] C O D E B A N C
[A D O B E C] O D E B A N C
CONTAINS ABC mark substring A D O B E C
...
[A D O B E C O D E B] A N C
B now > so moving left but A is compliance
...
[A D O B E C O D E B A] N C
A > moving left
A D O [B E C O D E B A] N C
B > moving left
A D O B E [C O D E B A] N C
C O D E B A is not shorter than A D O B E C
...
A D O B E [C O D E B A N C]
C > moving left
A D O B E C O D E [B A N C]
B A N C is shorter than A D O B E C mark substring B A N C
import java.util.HashMap;
import java.util.Map; public class MinimumWindowSubstring {
public void test(){
String s1 = "ADOBECODEBANC";
String t1 = "ABC";
// BANC
System.out.println(minWindow(s1,t1)); String s2 = "a";
String t2 = "aa";
// ""
System.out.println(minWindow(s2,t2)); String s3 = "a";
String t3 = "b";
// ""
System.out.println(minWindow(s3,t3)); String s4 = "a";
String t4 = "a";
// a
System.out.println(minWindow(s4,t4)); String s5 = "ab";
String t5 = "b";
// b
System.out.println(minWindow(s5,t5)); String s6 = "aa";
String t6 = "aa";
// aa
System.out.println(minWindow(s6,t6)); String s7 = "acbbaca";
String t7 = "aba";
// baca
System.out.println(minWindow(s7,t7)); String s8 = "aaaaaaaaaaaabbbbbcdd";
String t8 = "abcdd"; System.out.println(minWindow(s8,t8)); }
public String minWindow(String s, String t) {
if(s == null || t == null || s.isEmpty() || t.isEmpty() || s.length() < t.length()){
return "";
}
Map<Character,Integer> counter = new HashMap<>();
char[] sArray = s.toCharArray();
for(char c :t.toCharArray()){
counter.merge(c,1,(a,b)->a+b);
}
int left = 0;
int right ;
int[] ans = {-1 , 0 , 0};
Map<Character,Integer> window = new HashMap<>();
for(int i=0;i<sArray.length;i++){
char c = sArray[i];
right = i;
if(counter.containsKey(c)){
window.merge(c,1,(a,b)->a+b);
if(window.keySet().size() == counter.keySet().size()){
while (true){
Integer dec = window.get(sArray[left]);
Integer standard = counter.get(sArray[left]);
if(dec != null && standard != null){
if(dec <= standard){
break;
}
if(dec > standard){
window.merge(sArray[left],-1,(a,b)->a+b);
}
}
left ++;
}
boolean valid = true;
for(char v : counter.keySet()){
if(window.get(v) < counter.get(v)){
valid = false;
}
}
// mark if initial or shorter sequence
if(valid && (ans[0] == -1 || right - left + 1 < ans[0])){
ans[0] = right - left + 1;
ans[1] = left;
ans[2] = right;
}
}
}
}
return ans[0] == -1?"":s.substring(ans[1],ans[2]+1);
}
}
239. Sliding Window Maximum
Given an array nums, there is a sliding window of size k which is moving from the very left of the array to the very right. You can only see the k numbers in the window. Each time the sliding window moves right by one position. Return the max sliding window.
Input: nums = [,,-,-,,,,], and k =
Output: [,,,,,]
Explanation: Window position Max
--------------- -----
[ -] -
[ - -]
[- - ]
- [- ]
- - [ ]
- - [ ]
class Solution {
public int[] maxSlidingWindow(int[] nums, int k) {
List<Integer> result = new ArrayList<>();
if(nums == null || nums.length == 0){
return new int[]{};
}
int left = 0;
int windowMax = Integer.MIN_VALUE;
for(int i=0;i<nums.length;i++){
if(i - left + 1 > k){
int rv = nums[i];
int currentMax = windowMax;
if(nums[left] == windowMax){
currentMax = nums[left+1];
for(int t = left+1 ; t <= i ; t ++){
currentMax = Math.max(nums[t],currentMax);
}
}
windowMax = Math.max(currentMax,rv);
result.add(windowMax);
left ++ ;
} else {
windowMax = Math.max(nums[i],windowMax);
if(i - left + 1 == k){
result.add(windowMax);
}
}
}
int[] copyRes = new int[result.size()];
for(int index = 0 ; index < result.size() ; index ++){
copyRes[index] = result.get(index);
}
return copyRes;
}
}
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