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