LeetCode 567. Permutation in String C++

567. Permutation in String

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:
1. The input strings only contain lower case letters.
2. The length of both given strings is in range [1, 10,000].

Approach

1. 给你S1和S2字符串，问你S2是否包含S1的全排列中的一个。题目不难，一开始窝想到的是暴力解法，保存S1所有的全排列，然后去枚举S1的字符串进行匹配，然后我发现当字符串长度才10的时候全排列数量上千了，这样不仅爆内存还超时。然后我换另一种思路，他说全排列的一种可能，那么我就对S1进行映射，然后去枚举S1的字符串进行匹配，确实可行，不过速度有点慢60ms。

Code

class Solution {
public:
    bool isvalid(vector<int> let, string &str) {
        for (int i = 0; i < str.size(); i++) {
            int index = str[i] - 'a';
            let[index]--;
            if (let[index] < 0)return false;
        }
        return true;
    }
    bool checkInclusion(string s1, string s2) {
        vector<bool>letter(26, false);
        vector<int> let(26, 0);
        for (int i = 0; i < s1.size(); i++) {
            letter[s1[i] - 'a'] = true;
            let[s1[i] - 'a']++;
        }
        for (int i = 0; i < s2.size(); i++) {
            if (letter[s2[i] - 'a']) {
                string str = s2.substr(i, s1.size());
                if (str.size() < s1.size())return false;
                if (isvalid(let, str))return true;
            }
        }
        return false;
    }
};

Again

1. 看到某位大神的播客，深受启发，用滑动窗口的思想，维护S1长度的S2的子字符串，这样省去我上次代码提取字符串的过程，这样时间就大大缩短,而且代码也简洁。4ms

Code

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        vector<int>letters1(26, 0);
        vector<int>letters2(26, 0);
        for (int i = 0; i < s1.size(); i++) {
            letters1[s1[i] - 'a']++;
            letters2[s2[i] - 'a']++;
        }
        if (letters1 == letters2)return true;
        for (int i = s1.size(); i < s2.size(); i++) {
            letters2[s2[i] - 'a']++;
            letters2[s2[i - s1.size()] - 'a']--;
            if (letters1 == letters2)return true;
        }
        return false;
    }
};