# [567. Permutation in String](https://leetcode.com/problems/permutation-in-string/)

## 题目

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].

## 题目大意


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。

## 解题思路

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"。


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。








## 代码

```go

package leetcode

func checkInclusion(s1 string, s2 string) bool {
	var freq [256]int
	if len(s2) == 0 || len(s2) < len(s1) {
		return false
	}
	for i := 0; i < len(s1); i++ {
		freq[s1[i]-'a']++
	}
	left, right, count := 0, 0, len(s1)

	for right < len(s2) {
		if freq[s2[right]-'a'] >= 1 {
			count--
		}
		freq[s2[right]-'a']--
		right++
		if count == 0 {
			return true
		}
		if right-left == len(s1) {
			if freq[s2[left]-'a'] >= 0 {
				count++
			}
			freq[s2[left]-'a']++
			left++
		}
	}
	return false
}

```