Leetcode 567 Permutation in String (判断s1的全排列是否在s2中) (滑动窗口)

目录

• 问题描述
• 例子
• 方法一
• 方法二

Leetcode 567

问题描述

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.

例子

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

方法一

  采用固定的窗口，滑动一次检查一次。

** Solution Java **
** 4ms, beats 83.42% **
** 38.9MB, beats 19.23% **
class Solution {
    public boolean checkInclusion(String s1, String s2) {
        if (s2 == null || s2.length() < s1.length()) 
            return false;
        int[] count = new int[26];
        for (int i = 0; i < s1.length(); ++i) {
            ++count[s1.charAt(i) - 'a'];
            --count[s2.charAt(i) - 'a'];
        }
        if (countIsZero(count))
            return true;
        for (int j = s1.length(); j < s2.length(); ++j) {
            --count[s2.charAt(j) - 'a'];
            ++count[s2.charAt(j - s1.length()) - 'a'];
            if (countIsZero(count))
                return true;
        }
        return false;
    }
    private boolean countIsZero (int[] count) {
        for (int i = 0; i < 26; ++i)
            if (count[i] != 0)
                return false;
        return true;
    }
}

** Solution Python3 **
** 72ms, beats 62.30% **
** 12.9MB, beats 100.00% **
class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        if (s2 == None or len(s2) < len(s1)) :
            return False
        count = [0 for _ in range(26)]
        for i in range(len(s1)) :
            count[ord(s1[i]) - ord('a')] += 1
            count[ord(s2[i]) - ord('a')] -= 1
        if (self.isZero(count)) :
            return True
        for i in range(len(s1), len(s2)) :
            count[ord(s2[i]) - ord('a')] -= 1
            count[ord(s2[i - len(s1)]) - ord('a')] += 1
            if (self.isZero(count)) :
                return True
        return False
        
    def isZero(self, count):
        for i in range(26) :
            if (count[i] != 0) :
                return False
        return True

方法二

** Solution Java **
** 3ms, 99.10% **
** 39MB, 19.23% **
class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int[] count = new int[128];
        for (int i = 0; i < s1.length(); ++i)
            --count[s1.charAt(i)];
        for (int l = 0, r = 0; r < s2.length(); ++r) {
            if (++count[s2.charAt(r)] > 0)
                while (--count[s2.charAt(l++)] != 0) {}
            else if (r - l + 1 == s1.length())
                return true;
        }
        return s1.length() == 0;
    }
}