添加 problem 567

Author: halfrost

Algorithms/0567. Permutation in String/567. Permutation in String.go

@@ -0,0 +1,31 @@
+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
+}

Algorithms/0567. Permutation in String/567. Permutation in String_test.go

@@ -0,0 +1,63 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question567 struct {
+	para567
+	ans567
+}
+
+// para 是参数
+// one 代表第一个参数
+type para567 struct {
+	s string
+	p string
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans567 struct {
+	one bool
+}
+
+func Test_Problem567(t *testing.T) {
+
+	qs := []question567{
+
+		question567{
+			para567{"ab", "abab"},
+			ans567{true},
+		},
+
+		question567{
+			para567{"abc", "cbaebabacd"},
+			ans567{true},
+		},
+
+		question567{
+			para567{"abc", ""},
+			ans567{false},
+		},
+
+		question567{
+			para567{"abc", "abacbabc"},
+			ans567{true},
+		},
+
+		question567{
+			para567{"ab", "eidboaoo"},
+			ans567{false},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 567------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans567, q.para567
+		fmt.Printf("【input】:%v       【output】:%v\n", p, checkInclusion(p.s, p.p))
+	}
+	fmt.Printf("\n\n\n")
+}

Algorithms/0567. Permutation in String/README.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:
+
+```c
+Input:s1 = "ab" s2 = "eidbaooo"
+Output:True
+Explanation: s2 contains one permutation of s1 ("ba").
+```
+
+Example 2:
+
+```c
+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。
+
+
+
+
+
+