題目描述：

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".

要完成的函數：

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

說明：

1、給定一個字符串s和非空字符串p，將p中元素不斷交換形成一個新的字符串，如果這個新的字符串在s中能找到匹配的，那麽就輸出匹配到的開始的位置，直到處理到字符串s結束。

2、這道題目難道要記住p經過交換可能形成的所有字符串嗎，難道再類似於滑動窗口一般不斷在s中比較？

其實不用記住所有字符串，記住p經過交換可能形成的所有字符串其實等價於記住p中所有字母出現的次數。

所以代碼如下：

    vector<int> findAnagrams(string s, string p) 
    {
        vector<int>res;
        vector
 
<int>p1(26,0);
        vector<int>s1(26,0);
        for(int i=0;i<p.size();i++)
        {
            p1[p[i]-‘a‘]++;
        }
        for(int i=0;i<p.size();i++)
        {
            s1[s[i]-‘a‘]++;
        }
        if(s.size()<p.size())
            return res;
        if(p1==s1)
            res.push_back(0);
        for(int i=1;i<=s.size()-p.size();i++)
        {
            s1[s[i-1]-‘a‘]--;//當窗口滑動時，s1中要減去窗口前一位的字母1次
            s1[s[i-1+p.size()]-‘a‘]++;//s1要加上窗口最後新的一維的字母1次
            if(s1==p1)
                res.push_back(i);
        }
        return res;
    }

上述代碼實測35ms，beats 90.84% of cpp submissions。

leetcode-438-Find All Anagrams in a String