567字符串排列

这种思路就是固定窗口大小，然后不断向后滑动，直到找到一个排列相等的字符串 

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

这种思路就是双指针滑动窗口，快指针先走，达到一定条件就缩短窗口，反复执行 

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        vector<int>need(26,0);
        for(auto ch:s1)
            need[ch-'a']++;
        int fast=0,slow=0;
        while(fast<s2.size()){
            need[s2[fast]-'a']--;
            while(need[s2[fast]-'a']<0){
                need[s2[slow]-'a']++;
                slow++;
            }
            if(fast-slow+1==s1.size()){
                return true;
            }
            fast++;
        }
        return false;
    }
};