LeetCode : Find All Anagrams in a String

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”.
http://blog.csdn.net/shiyang6017/article/details/52901357

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> cnt(27, 0);
        for (char ch : p)
            cnt[ch - 'a']++;
        vector<int> ret;
        int start = 0;
        s += '{'; // 哨兵  
        /*
        * 若不添加哨兵，代码量会稍微大些，比较麻烦
        * 考虑不添加哨兵，若s[i, s.size() - 1]满足条件时，下面for循环会溜掉这种情况。
        */
        for (int i = 0; i < s.size(); ++i)
        if (--cnt[s[i] - 'a'] < 0){
            if (i - start == p.size())
                ret.push_back(start);
            while (cnt[s[i] - 'a'] < 0 && start < s.size())
                ++cnt[s[start++] - 'a'];
        }
        return ret;
    }
};

class Solution {
public:
    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;
    }
};