题目链接：https://leetcode.com/problems/permutation-in-string/#/description
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:
The input strings only contain lower case letters.The length of both given strings is in range [1, 10,000].思路：观察字符串s1和s2，如果s1的全排列是s2的子串，则s1的所有字符均出现在s2中，而且这些字符连续。
于是用一个类似于滑动窗口的区间去统计此区间内的字符，区间的长度取为s1的长度。
如果在s2中存在一个区间，里面的字符与s1的字符全部都一样，则说明s1通过全排列后一定能够得到s2中的子串。也即满足题意
class Solution{ public: bool checkInclusion(string s1, string s2) { int m=s1.length(),n=s2.length(); if(m>n) return false; vector<int>maps1(26),maps2(26); for(int i=0;i<m;i++) { maps1[s1[i]-'a']++; maps2[s2[i]-'a']++; } if(maps1==maps2) return true; for(int i=0;i+m<n;i++) { maps2[s2[i]-'a']--; maps2[s2[i+m]-'a']++; if(maps1==maps2) return true; } return false; } };
