chore:

Author: kimi0230

Leetcode/0567.Permutation-in-String/main.go

@@ -81,3 +81,38 @@ func CheckInclusionSlice(s1 string, s2 string) bool {
 	}
 	return false
 }
+
+func CheckInclusion2(s1 string, s2 string) bool {
+	left, right := 0, 0
+	need, windows := make(map[byte]int), make(map[byte]int)
+	match := 0
+	for _, v := range s1 {
+		need[byte(v)]++
+	}
+
+	for right < len(s2) {
+		cRight := s2[right]
+		windows[cRight]++
+		right++
+		if windows[cRight] == need[cRight] {
+			match++
+		}
+
+		if match == len(need) {
+			return true
+		}
+
+		if right-left >= len(s1) {
+			cLeft := s2[left]
+
+			// 如果左指針字符出現次數達到目標，減少 match
+			if windows[cLeft] == need[cLeft] {
+				match--
+			}
+			// 移動左指針
+			windows[cLeft]--
+			left++
+		}
+	}
+	return false
+}

Leetcode/0567.Permutation-in-String/main_test.go

@@ -7,16 +7,21 @@ var tests = []struct {
 	arg2 string
 	want bool
 }{
+	// {
+	// 	"ab",
+	// 	"eidbaooo",
+	// 	true,
+	// },
+	// {
+	// 	"ab",
+	// 	"eidboaoo",
+	// 	false,
+	// },
 	{
-		"ab",
-		"eidbaooo",
+		"abcdxabcde",
+		"abcdeabcdx",
 		true,
 	},
-	{
-		"ab",
-		"eidboaoo",
-		false,
-	},
 }
 
 func TestCheckInclusion(t *testing.T) {
@@ -35,6 +40,14 @@ func TestCheckInclusionSlice(t *testing.T) {
 	}
 }
 
+func TestCheckInclusion2(t *testing.T) {
+	for _, tt := range tests {
+		if got := CheckInclusion2(tt.arg1, tt.arg2); got != tt.want {
+			t.Errorf("got = %v, want = %v", got, tt.want)
+		}
+	}
+}
+
 func BenchmarkCheckInclusion(b *testing.B) {
 	b.ResetTimer()
 	for i := 0; i < b.N; i++ {

Leetcode_labuladong/567.字符串的排列.go

@@ -0,0 +1,87 @@
+// 時間複雜度:
+// 空間複雜度:
+/*
+ * @lc app=leetcode.cn id=567 lang=golang
+ *
+ * [567] 字符串的排列
+ *
+ * https://leetcode.cn/problems/permutation-in-string/description/
+ *
+ * algorithms
+ * Medium (44.95%)
+ * Likes:    998
+ * Dislikes: 0
+ * Total Accepted:    287.5K
+ * Total Submissions: 639.1K
+ * Testcase Example:  '"ab"\n"eidbaooo"'
+ *
+ * 给你两个字符串 s1 和 s2 ，写一个函数来判断 s2 是否包含 s1 的排列。如果是，返回 true ；否则，返回 false 。
+ *
+ * 换句话说，s1 的排列之一是 s2 的 子串 。
+ *
+ *
+ *
+ * 示例 1：
+ *
+ *
+ * 输入：s1 = "ab" s2 = "eidbaooo"
+ * 输出：true
+ * 解释：s2 包含 s1 的排列之一 ("ba").
+ *
+ *
+ * 示例 2：
+ *
+ *
+ * 输入：s1= "ab" s2 = "eidboaoo"
+ * 输出：false
+ *
+ *
+ *
+ *
+ * 提示：
+ *
+ *
+ * 1 <= s1.length, s2.length <= 10^4
+ * s1 和 s2 仅包含小写字母
+ *
+ *
+ */
+
+// @lc code=start
+func checkInclusion(s1 string, s2 string) bool {
+	left, right := 0, 0
+	need, windows := make(map[byte]int), make(map[byte]int)
+	match := 0
+	for _, v := range s1 {
+		need[byte(v)]++
+	}
+
+	for right < len(s2) {
+		cRight := s2[right]
+		windows[cRight]++
+		right++
+		if windows[cRight] == need[cRight] {
+			match++
+		}
+
+		if match == len(need) {
+			return true
+		}
+
+		if right-left >= len(s1) {
+			cLeft := s2[left]
+
+			// 如果左指針字符出現次數達到目標，減少 match
+			if windows[cLeft] == need[cLeft] {
+				match--
+			}
+			// 移動左指針
+			windows[cLeft]--
+			left++
+		}
+	}
+	return false
+}
+
+// @lc code=end
+