567 Permutation in String

最新推荐文章于 2018-10-29 09:58:00 发布

原创最新推荐文章于 2018-10-29 09:58:00 发布 · 189 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 同时被 2 个专栏收录

146 篇文章

订阅专栏

MicroSoft

85 篇文章

订阅专栏

本文介绍了一种使用滑动窗口的方法来判断一个字符串是否为另一个字符串的排列。通过比较两个字符串中字符出现的频率，可以在O(n)的时间复杂度内完成判断。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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].

使用滑动窗口Sliding Window

class Solution {
    public boolean checkInclusion(String s1, String s2) {
        if (s1.length() > s2.length()) {
            return false;
        }
        final int ASCII = 256;
        int[] m1 = new int[ASCII];
        int[] m2 = new int[ASCII];
        
        for (int i = 0; i < s1.length(); i++) {
            m1[s1.charAt(i)]++;
            m2[s2.charAt(i)]++;
        }
        if (Arrays.equals(m1, m2)) {
            return true;
        }
        for (int i = s1.length(); i < s2.length(); i++) {
            m2[s2.charAt(i)]++;
            m2[s2.charAt(i - s1.length())]--;
            if (Arrays.equals(m1, m2)) {
                return true;
            }
        }
        return false;
    }
}