Minimum Window Substring &&& Longest Substring Without Repeating Characters 快慢指针，都不会退，用hashmap或者其他结构保证

 1 

public class Solution {
    public static int lengthOfLongestSubstring(String s) {
 
    char[] arr = s.toCharArray();
    int pre = 0;
 
    HashMap<Character, Integer> map = new HashMap<Character, Integer>();
 
    for (int i = 0; i < arr.length; i++) {
        if (!map.containsKey(arr[i])) {
            map.put(arr[i], i);
        } else {
            pre = pre > map.size() ? pre : map.size();
            i = map.get(arr[i]);
            map.clear();
        }
    }
 
    return Math.max(pre, map.size());
}
}

public class Solution {
    public String minWindow(String S, String T) {
        char s[]=S.toCharArray();
        if(S=="")return "";
        int beg=0;
        int end=0;
        int d[]=new  int[128];
        int size=0;
        for(int i=0;i<T.length();i++)
        {
            if(d[t[i]]==0) size++;
            d[t[i]]++;
        }
        int d2[]=new int[128];
        int s1=0;
        boolean flag=false;
        
        while(end<s.length)
        {
            if(d[s[end]]==0) {end++;continue;}
             d2[s[end]]++; 
             if(d2[s[end]]==d[s[end]]) s1++;
             end++;
             if(s1==size)
             {
                 while(d2[s[beg]]>d[s[beg]]||d[s[beg]]==0) {d2[s[beg]]--;beg++;}
                 flag=true;
                 break;
             }
           
            
        }
       if(!flag) return "";
        int aend=end-1;
        int abeg=beg;
        int amin=end-beg;
        
    
         while(end<s.length)
         {
             if(d[s[end]]==0){end++;continue;}
             d2[s[end]]++;
            
            while((d2[s[beg]]>d[s[beg]])||d[s[beg]]==0) { 
                d2[s[beg]]--;beg++;
                if(end-beg+1<amin)
                {amin=end-beg+1;
                aend=end;
                abeg=beg;
            
                }
            
                
                
                
            }
            end++;
             
         }
        
            
        
        // 
        return S.substring(abeg,aend+1);
        
        
        
        
        
        
      
        
    }
}