76. Minimum Window Substring（hard 双指针）

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"

Note:

If there is no such window in S that covers all characters in T, return the empty string "".
If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

 1 class Solution {
 2     public String minWindow(String s, String t) {
 3         
 4         if(s.length()<t.length())
 5             return "";
 6         Map<Character,Integer> wordDict = constructWordDict(t);
 7   
 8         int slow =0,minLen=Integer.MAX_VALUE,fast=0,matchCount=0,start = 0;
 9         for(fast=0;fast<s.length();fast++){
10             char ch = s.charAt(fast);
11             Integer cnt = wordDict.get(ch);
12             
13             //当前字符不在map中，fast++
14             if(cnt ==null)
15                 continue;
16             wordDict.put(ch,cnt-1);
17             
18             //当前字符在map中，且 cnt==1，需要这个match,  match总数++
19             if(cnt==1)
20                 matchCount++;
21             
22             // 如果 match的个数够了，尝试移动slow,使其更短
23             while(matchCount==wordDict.size()){
24                 //更新最短长度
25                 if(fast-slow+1<minLen){
26                     minLen = fast-slow+1;
27                     start=slow;
28                 }
29                 //移动slow
30                 char left = s.charAt(slow++);
31                 Integer leftCnt = wordDict.get(left);
32                 //当 slow 对应的字符串不在map中，说明当前字符串不需要match，继续移动
33                 if(leftCnt==null)
34                     continue;
35                 
36                 //当slow 对应的字符串在map中时，map中的key+1,
37                 wordDict.put(left,leftCnt+1);
38                 //如果slow对应的元素cnt==0，说明移动过头了，需要重新match slow对应的元素
39                 if(leftCnt==0)
40                     matchCount --;
41                 
42             }
43         }
44         return minLen==Integer.MAX_VALUE?"":s.substring(start,start+minLen);
45     }
46     private Map<Character,Integer> constructWordDict(String s){
47         Map<Character,Integer> map = new HashMap<>();
48         for(char ch :s.toCharArray()){
49             Integer cnt = map.get(ch);
50             if(cnt==null)
51                 map.put(ch,1);
52             else
53                 map.put(ch,cnt+1);
54         }
55         return map;
56     }
57 }

https://www.youtube.com/watch?v=9qFR2WQGqkU