Question:

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".
 

Algorithm:

Accepted Code：

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> res;
        if(p.size()>s.size())
            return res;
        vector<int> hash1(26,0);
        vector<int> hash2(26,0);
        for(int i=0;i<p.size();i++)
        {
            hash1[p[i]-'a']++;
            hash2[s[i]-'a']++;
        }
        if(hash1==hash2)
            res.push_back(0);
        for(int i=p.size();i<s.size();i++)
        {
            hash2[s[i]-'a']++;
            hash2[s[i-p.size()]-'a']--;
            if(hash2==hash1)
                res.push_back(i-p.size()+1);
        }
        return res;
    }
};