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".

思路：

用一個滑動窗口，隨時判斷窗口裏面的字符是否與p字符串相符合anagram。

vector<int> findAnagrams(string s, string p) {
        vector<int> pv(256,0), sv(256,0), res;
        if(s.size() < p.size())
           return res;
        for(int i = 0; i < p.size(); ++i)
        {
            ++pv[p[i]];
            ++sv[s[i]];
        }
        if(pv == sv)
           res.push_back(
 
0);
        for(int i = p.size(); i < s.size(); ++i)
        {
            ++sv[s[i]];
            --sv[s[i-p.size()]];
            if(pv == sv)
               res.push_back(i-p.size()+1);
        }
        return res;
    }

參考：

https://discuss.leetcode.com/topic/64390/c-o-n-sliding-window-concise-solution-with-explanation

[leetcode-438-Find All Anagrams in a String]