原题链接在这里：

题目：

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:s: "cbaebabacd" p: "abc"Output:[0, 6]Explanation:The substring with start index = 0 is "cba", which is an anagram of "abc".The substring with start index = 6 is "bac", which is an anagram of "abc".

Example 2:

Input:s: "abab" p: "ab"Output:[0, 1, 2]Explanation:The substring with start index = 0 is "ab", which is an anagram of "ab".The substring with start index = 1 is "ba", which is an anagram of "ab".The substring with start index = 2 is "ab", which is an anagram of "ab". 

题解：

先将p的所有char对应的map count++.

然后快慢指针维护一个sliding window. 右侧runner每次都移动，对应的char的count减1. 若是count减1之前大于0, 表示char在p中, sum--.

sum为0时就找到了anagram.

当window大小为p长度时，移动左侧walker, 对应char的count加1. 若是count加1之前不是负数，表示这个char在p中, sum++.

Time Complexity: O(n). n = s.length().

Space: O(1). 用到了map.

AC Java:

1 public class Solution { 2     public List
     
       findAnagrams(String s, String p) { 3         List
      
        res = new ArrayList
       
        (); 4         if(s == null || p == null || s.length() == 0 || p.length() == 0){ 5             return res; 6         } 7         int [] map = new int[256]; 8         for(int i = 0; i
        
          0){16                 //Move runner everytime. Decrease corresponding char count by 117                 //If current char count is larger than 0, then this char is in p18                 sum--;19             }20             21             if(sum == 0) {22                 //Find anagram23                 res.add(walker);24             }25 26             if(runner-walker == p.length() && map[s.charAt(walker++)]++ >= 0){27                 //If windows's size is already p.length(), move walker and increase corresponding char count by 1 28                 //If count before increasing is not smaller than 0, this char is in p29                 sum++;30             }31         }32         return res;33     }34 }