feat: leetcode 面试题 01.02

Author: deweyou

README.md

@@ -2,7 +2,7 @@
 
 <p align="center">
 <!-- TOPICS COUNT START -->
-<img src="https://img.shields.io/badge/-进度:505-green" alt="进度:505">
+<img src="https://img.shields.io/badge/-进度:506-green" alt="进度:506">
 <!-- TOPICS COUNT END -->
 <a href="./assets/docs/TOPICS.md"><img src="https://img.shields.io/badge/-题库目录-blue" alt="题库目录"></a>
 <a href="./assets/docs/CATEGORIES.md"><img src="https://img.shields.io/badge/-题库分类-red" alt="题库分类"></a>

assets/data/categories.json

@@ -361,6 +361,12 @@
         "title": "检测正方形",
         "path": "./problemset/detect-squares/README.md",
         "difficulty": "中等"
+      },
+      {
+        "id": "面试题 01.02",
+        "title": "判定是否互为字符重排",
+        "path": "./problemset/check-permutation-lcci/README.md",
+        "difficulty": "简单"
       }
     ]
   },

assets/data/topics.json

@@ -5048,5 +5048,15 @@
     "difficulty": "困难",
     "url": "https://leetcode.cn/problems/missing-two-lcci/",
     "path": "./problemset/missing-two-lcci/README.md"
+  },
+  {
+    "id": "面试题 01.02",
+    "title": {
+      "cn": "判定是否互为字符重排",
+      "en": "check-permutation-lcci"
+    },
+    "difficulty": "简单",
+    "url": "https://leetcode.cn/problems/check-permutation-lcci/",
+    "path": "./problemset/check-permutation-lcci/README.md"
   }
 ]

assets/docs/CATEGORIES.md

@@ -64,6 +64,7 @@
 | 1832. [判断句子是否为全字母句](../../problemset/check-if-the-sentence-is-pangram/README.md) | 简单 |
 | 2006. [差的绝对值为 K 的数对数目](../../problemset/count-number-of-pairs-with-absolute-difference-k/README.md) | 简单 |
 | 2013. [检测正方形](../../problemset/detect-squares/README.md) | 中等 |
+| 面试题 01.02. [判定是否互为字符重排](../../problemset/check-permutation-lcci/README.md) | 简单 |
 
 ## 链表
 

assets/docs/TOPICS.md

@@ -998,8 +998,6 @@
 
 [面试题 17.11. 单词距离](../../problemset/find-closest-lcci/README.md)
 
-[面试题 17.19. 消失的两个数字](../../problemset/missing-two-lcci/README.md)
-
 [剑指 Offer II 114. 外星文字典](../../problemset/alien-dictionary/README.md)
 
 [剑指 Offer II 029. 排序的循环链表](../../problemset/sorted-circular-linked-list/README.md)
@@ -1008,4 +1006,8 @@
 
 [剑指 Offer II 041. 滑动窗口的平均值](../../problemset/average-value-of-sliding-window/README.md)
 
-[剑指 Offer II 115. 重建序列](../../problemset/rebuild-sequence/README.md)
+[剑指 Offer II 115. 重建序列](../../problemset/rebuild-sequence/README.md)
+
+[面试题 17.19. 消失的两个数字](../../problemset/missing-two-lcci/README.md)
+
+[面试题 01.02. 判定是否互为字符重排](../../problemset/check-permutation-lcci/README.md)

problemset/check-permutation-lcci/README.md

@@ -0,0 +1,50 @@
+# 判定是否互为字符重排
+
+> 难度：简单
+>
+> https://leetcode.cn/problems/check-permutation-lcci/
+
+## 题目
+
+给定两个字符串 `s1` 和 `s2`，请编写一个程序，确定其中一个字符串的字符重新排列后，能否变成另一个字符串。
+
+### 示例
+
+#### 示例 1：
+
+```
+输入: s1 = "abc", s2 = "bca"
+输出: true 
+```
+
+#### 示例 2：
+
+```
+输入: s1 = "abc", s2 = "bad"
+输出: false
+```
+
+## 解题
+
+```ts 
+/**
+ * 时间复杂度 O(N)  空间复杂度 O(N)
+ * @param s1
+ * @param s2
+ * @returns
+ */
+export function CheckPermutation(s1: string, s2: string): boolean {
+  const map = new Map<string, number>()
+  for (const s of s1)
+    map.set(s, (map.get(s) || 0) + 1)
+
+  for (const s of s2) {
+    const count = map.get(s)!
+    if (!count) return false
+    if (count === 1) map.delete(s)
+    else map.set(s, count - 1)
+  }
+
+  return map.size === 0
+}
+```

problemset/check-permutation-lcci/index.spec.ts

@@ -0,0 +1,19 @@
+import { describe, expect, it } from 'vitest'
+import { CheckPermutation } from '.'
+
+describe('判定是否互为字符重排', () => {
+  testCase(CheckPermutation)
+})
+
+function testCase(fn: (s1: string, s2: string) => boolean) {
+  it.each([
+    [
+      'abc', 'bca', true,
+    ],
+    [
+      'abc', 'bad', false,
+    ],
+  ])('示例%#', (s1, s2, expected) => {
+    expect(fn(s1, s2)).toBe(expected)
+  })
+}

problemset/check-permutation-lcci/index.ts

@@ -0,0 +1,20 @@
+/**
+ * 时间复杂度 O(N)  空间复杂度 O(N)
+ * @param s1
+ * @param s2
+ * @returns
+ */
+export function CheckPermutation(s1: string, s2: string): boolean {
+  const map = new Map<string, number>()
+  for (const s of s1)
+    map.set(s, (map.get(s) || 0) + 1)
+
+  for (const s of s2) {
+    const count = map.get(s)!
+    if (!count) return false
+    if (count === 1) map.delete(s)
+    else map.set(s, count - 1)
+  }
+
+  return map.size === 0
+}