567.字符串的排列(滑动窗口+双指针) 难度：中等 语言：C++

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        int pFreq[26];
        int winFreq[26]; //记录s2中字符出现频率

        int right = 0;
        int left = 0;

        int pcount = 0; //s1中字符种类数
        int wincount = 0; //s2中字符种类数（只有当s2中该字符出现频率与s1中该字符出现频率相同时，自加）

        for(int i = 0; i < 26; i++){
            pFreq[i] = 0;
            winFreq[i] = 0;
        }

        for(int i = 0; i < s1.length(); i++){
            pFreq[s1[i]-'a']++; //存储s1中每个字符出现的频率
        }

        for(int i = 0; i < 26; i++){
            if(pFreq[i] > 0)
                pcount++; //存储出现的字符种类数
        }

        while(right < s2.length()){
            if(pFreq[s2[right]-'a'] > 0){ //如果s2右指针指向的字符存在于s1中
                winFreq[s2[right]-'a']++; //winFreq数组中该字符数加1
                if(pFreq[s2[right]-'a'] == winFreq[s2[right]-'a']){ //如果s2中该字符出现频率与s1相同
                    wincount++; //s2字符种类数加1
                }
            }
            right++; //指针右移

            while(pcount == wincount){ //如果s1与s2字符种类数相同
                if(right - left == s1.length()) //且s2子串长度与s1相同
                    return true; //说明找到满足条件字符串
                if(pFreq[s2[left]-'a'] > 0){ //如果左指针指向的字符存在于s1中
                    winFreq[s2[left]-'a']--; //winFreq数组中该字符减1(因为左指针左移，剔除该字符)
                    if(winFreq[s2[left]-'a'] < pFreq[s2[left]-'a']){ //如果s2中该字符出现频率小于s1
                        wincount--; //s2字符种类数减1
                    }
                }
                left++; //左指针左移
            }
        }
        return false;
    }
};