Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string‘s permutations is the substring of the second string.

Example 1:

Input:s1 = "ab" s2 = "eidbaooo"
Output:True
Explanation: s2 contains one permutation of s1 ("ba").

Example 2:

Input:s1= "ab" s2 = "eidboaoo"
Output: False

Note:

1. The input strings only contain lower case letters.
2. The length of both given strings is in range [1, 10,000].

維護一下一段區間的字符串就行，有O(26*N)的做法，我這裏是用multiset做得，時間復雜度理論上 不必O(26*n)大啊但是跑的慢 可能是判斷集合是否相等的時候比較費時間 有時間好好看看multiset源碼

class
 
 Solution {
public:
    bool checkInclusion(string s1, string s2) {
        multiset<char>se1,se2;
        if (s2.size() < s1.size()) return false;
        for (int i = 0; i < s1.size(); ++i) se1.insert(s1[i]),se2.insert(s2[i]);
        if (se2 == se1) return true;
       
        
 
for (int i = s1.size(); i < s2.size(); ++i) {
            char c = s2[i - s1.size()];
            auto x = se2.find(c);
            se2.erase(x);
            se2.insert(s2[i]);
            if (se1 == se2) return true;
        }
        return false;
    }
};

leetcode 567. Permutation in String