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87 lines
1.6 KiB
Markdown
87 lines
1.6 KiB
Markdown
# [567. Permutation in String](https://leetcode.com/problems/permutation-in-string/)
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## 题目
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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.
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**Example 1**:
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```
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Input:s1 = "ab" s2 = "eidbaooo"
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Output:True
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Explanation: s2 contains one permutation of s1 ("ba").
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```
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**Example 2**:
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```
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Input:s1= "ab" s2 = "eidboaoo"
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Output: False
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```
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**Note**:
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1. The input strings only contain lower case letters.
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2. The length of both given strings is in range [1, 10,000].
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## 题目大意
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在一个字符串重寻找子串出现的位置。子串可以是 Anagrams 形式存在的。Anagrams 是一个字符串任意字符的全排列组合。
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## 解题思路
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这一题和第 438 题，第 3 题，第 76 题，第 567 题类似，用的思想都是"滑动窗口"。
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这道题只需要判断是否存在，而不需要输出子串所在的下标起始位置。所以这道题是第 438 题的缩水版。具体解题思路见第 438 题。
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## 代码
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```go
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package leetcode
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func checkInclusion(s1 string, s2 string) bool {
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var freq [256]int
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if len(s2) == 0 || len(s2) < len(s1) {
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return false
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}
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for i := 0; i < len(s1); i++ {
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freq[s1[i]-'a']++
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}
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left, right, count := 0, 0, len(s1)
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for right < len(s2) {
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if freq[s2[right]-'a'] >= 1 {
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count--
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}
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freq[s2[right]-'a']--
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right++
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if count == 0 {
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return true
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}
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if right-left == len(s1) {
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if freq[s2[left]-'a'] >= 0 {
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count++
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}
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freq[s2[left]-'a']++
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left++
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}
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}
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return false
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}
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``` |