LeetCode567. Permutation in String

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本文介绍了一种检查一个字符串是否为另一个字符串的排列的方法。通过滑动窗口比较两个字符串中字符出现的频率来判断是否存在子串是第一个字符串的排列。

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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].

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        int n1=s1.length(),n2=s2.length();
        if(n2<n1) return false;
        if(n1==0) return true;
        int i=0,j=0;
        vector<int> v1(26),v2(26);
        for(i=0;i<n1;++i){
            v1[s1[i]-'a']++;
            v2[s2[i]-'a']++;
        }
        if(v1==v2) return true;
        for(j=n1;j<n2;++j){
           v2[s2[j-n1]-'a']--;
           v2[s2[j]-'a']++;
           if(v1==v2) return true;
        }
        return false;
    }
};