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.

class Solution {
    public String minWindow(String s, String t) {
        
        if(s.length()<t.length())
            return "";
        Map<Character,Integer> wordDict = constructWordDict(t);
  
        int slow =0,minLen=Integer.MAX_VALUE,fast=0,matchCount=0,start = 0;
        for(fast=0;fast<s.length();fast++){
            char ch = s.charAt(fast);
            Integer cnt = wordDict.get(ch);
            
            //当前字符不在map中，fast++
            if(cnt ==null)
                continue;
            wordDict.put(ch,cnt-1);
            
            //当前字符在map中，且 cnt==1，需要这个match,  match总数++
            if(cnt==1)
                matchCount++;
            
            // 如果 match的个数够了，尝试移动slow,使其更短
            while(matchCount==wordDict.size()){
                //更新最短长度
                if(fast-slow+1<minLen){
                    minLen = fast-slow+1;
                    start=slow;
                }
                //移动slow
                char left = s.charAt(slow++);
                Integer leftCnt = wordDict.get(left);
                //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
                if(leftCnt==null)
                    continue;
                
                //当slow 对应的字符串在map中时，map中的key+1,
                wordDict.put(left,leftCnt+1);
                //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
                if(leftCnt==0)
                    matchCount --;
                
            }
        }
        return minLen==Integer.MAX_VALUE?"":s.substring(start,start+minLen);
    }
    private Map<Character,Integer> constructWordDict(String s){
        Map<Character,Integer> map = new HashMap<>();
        for(char ch :s.toCharArray()){
            Integer cnt = map.get(ch);
            if(cnt==null)
                map.put(ch,1);
            else
                map.put(ch,cnt+1);
        }
        return map;
    }
}

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