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website/content/ChapterFour/0567.Permutation-in-String.md

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+# [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
+}
+
+```