【LeetCode】C++ 567. Permutation in String

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该博客讨论了如何使用C++解决LeetCode上的567题，即判断一个字符串是否包含另一个字符串的排列作为子串。通过分析字符串的字符出现频率，可以有效地进行判断。

Given two strings s1 and s2, return true if s2 contains a permutation of s1, or false otherwise.

In other words, return true if one of s1's permutations is the substring of s2.

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

Constraints:

1 <= s1.length, s2.length <= 104
s1 and s2 consist of lowercase English letters.

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        if (s1.size() > s2.size()) return false;
        vector<int> c1(26),c2(26);
        for (int i=0;i<s1.size();i++){
            ++c1[s1[i]-'a'];
            ++c2[s2[i]-'a'];
        }
        for (int i=s1.size(),j=s1.size(); i<s2.size();i++){
            if (c1==c2) return true;
            --c2[s2[i-j]-'a'];
            ++c2[s2[i]-'a'];
        }
        return c1==c2;

    }
};