【LeetCode】567. Permutation in String

567. Permutation in String

介绍

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:TrueExplanation: 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].

解答

class Solution {public:    //在字符串s2上设置一个滑动窗口,窗口大小是s1的长度,    //如果在这个窗口内包含s1的所有字母元素,    //就说明这个窗口内的所有元素组合在一起之后一定是s1的一个全排列    bool checkInclusion(string s1, string s2)     {        int len1 = s1.size(),len2 = s2.size();        if(len2 < len1) return false;        if(len1 == 0)   return true;        vector<int> mapping1(26),mapping2(26);        for(int i = 0; i < len1; ++i)        {            ++mapping1[s1[i]-'a'];            ++mapping2[s2[i]-'a'];        }        if(mapping1 == mapping2)    return true;        for(int i = 1; i+len1-1 < len2; ++i)        {            --mapping2[s2[i-1]-'a'];            ++mapping2[s2[i+len1-1]-'a'];            if(mapping1 == mapping2)                return true;        }        return false;    }};