题目描述：

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".

要完成的函数：

vector<int> findAnagrams(string s, string p) 

说明：

1、给定一个字符串s和非空字符串p，将p中元素不断交换形成一个新的字符串，如果这个新的字符串在s中能找到匹配的，那么就输出匹配到的开始的位置，直到处理到字符串s结束。

2、这道题目难道要记住p经过交换可能形成的所有字符串吗，难道再类似于滑动窗口一般不断在s中比较？

其实不用记住所有字符串，记住p经过交换可能形成的所有字符串其实等价于记住p中所有字母出现的次数。

所以代码如下：

vector
     
       findAnagrams(string s, string p)     {        vector
      
       res;        vector
       
        p1(26,0);        vector
        
         s1(26,0);        for(int i=0;i
        
       
      
     

上述代码实测35ms，beats 90.84% of cpp submissions。