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);

     }
 }