feat: leetcode 面试题 01.09

Author: ouduidui

README.md

@@ -2,7 +2,7 @@
 
 <p align="center">
 <!-- TOPICS COUNT START -->
-<img src="https://img.shields.io/badge/-进度:507-green" alt="进度:507">
+<img src="https://img.shields.io/badge/-进度:508-green" alt="进度:508">
 <!-- 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

@@ -2566,6 +2566,12 @@
         "title": "滑动窗口的平均值",
         "path": "./problemset/average-value-of-sliding-window/README.md",
         "difficulty": "简单"
+      },
+      {
+        "id": "面试题 01.09",
+        "title": "字符串轮转",
+        "path": "./problemset/string-rotation-lcci/README.md",
+        "difficulty": "简单"
       }
     ]
   },

assets/data/topics.json

@@ -5068,5 +5068,15 @@
     "difficulty": "中等",
     "url": "https://leetcode.cn/problems/get-kth-magic-number-lcci/",
     "path": "./problemset/get-kth-magic-number-lcci/README.md"
+  },
+  {
+    "id": "面试题 01.09",
+    "title": {
+      "cn": "字符串轮转",
+      "en": "string-rotation-lcci"
+    },
+    "difficulty": "简单",
+    "url": "https://leetcode.cn/problems/string-rotation-lcci/",
+    "path": "./problemset/string-rotation-lcci/README.md"
   }
 ]

assets/docs/CATEGORIES.md

@@ -469,6 +469,7 @@
 | 面试题 17.11. [单词距离](../../problemset/find-closest-lcci/README.md) | 中等 |
 | 剑指 Offer II 029. [排序的循环链表](../../problemset/sorted-circular-linked-list/README.md) | 中等 |
 | 剑指 Offer II 041. [滑动窗口的平均值](../../problemset/average-value-of-sliding-window/README.md) | 简单 |
+| 面试题 01.09. [字符串轮转](../../problemset/string-rotation-lcci/README.md) | 简单 |
 
 ## 回溯
 

assets/docs/TOPICS.md

@@ -1012,4 +1012,6 @@
 
 [面试题 01.02. 判定是否互为字符重排](../../problemset/check-permutation-lcci/README.md)
 
-[面试题 17.09. 第 k 个数](../../problemset/get-kth-magic-number-lcci/README.md)
+[面试题 17.09. 第 k 个数](../../problemset/get-kth-magic-number-lcci/README.md)
+
+[面试题 01.09. 字符串轮转](../../problemset/string-rotation-lcci/README.md)

problemset/string-rotation-lcci/README.md

@@ -0,0 +1,72 @@
+# 字符串轮转
+
+> 难度：简单
+>
+> https://leetcode.cn/problems/string-rotation-lcci/
+
+## 题目
+
+字符串轮转。给定两个字符串s1和s2，请编写代码检查s2是否为s1旋转而成（比如，waterbottle是erbottlewat旋转后的字符串）。
+
+### 示例
+
+#### 示例1:
+
+```
+ 输入：s1 = "waterbottle", s2 = "erbottlewat"
+ 输出：True
+```
+
+#### 示例2:
+
+```
+ 输入：s1 = "aa", s2 = "aba"
+ 输出：False
+```
+
+## 解题
+
+### 模拟
+
+```ts 
+/**
+ * 模拟
+ * @desc 时间复杂度 O(N²)  空间复杂度 O(1)
+ * @param s1
+ * @param s2
+ * @returns
+ */
+export function isFlipedString(s1: string, s2: string): boolean {
+  const m = s1.length
+  const n = s2.length
+  if (m !== n) return false
+  if (n === 0) return true
+
+  for (let i = 0; i < n; i++) {
+    let flag = true
+    for (let j = 0; j < n; j++) {
+      if (s1[(i + j) % n] !== s2[j]) {
+        flag = false
+        break
+      }
+    }
+    if (flag) return true
+  }
+  return false
+}
+```
+
+### 搜索子字符串
+
+```ts 
+/**
+ * 搜索子字符串
+ * @desc 时间复杂度 O(N)  空间复杂度 O(N)
+ * @param s1
+ * @param s2
+ * @returns
+ */
+export function isFlipedString2(s1: string, s2: string): boolean {
+  return s1.length === s2.length && (s1 + s1).includes(s2)
+}
+```

problemset/string-rotation-lcci/index.spec.ts

@@ -0,0 +1,16 @@
+import { describe, expect, it } from 'vitest'
+import { isFlipedString, isFlipedString2 } from '.'
+
+describe('字符串轮转', () => {
+  describe('模拟', () => testCase(isFlipedString))
+  describe('搜索子字符串', () => testCase(isFlipedString2))
+})
+
+function testCase(fn: (s1: string, s2: string) => boolean) {
+  it.each([
+    ['waterbottle', 'erbottlewat', true],
+    ['aa', 'aba', false],
+  ])('示例%#', (s1, s2, expected) => {
+    expect(fn(s1, s2)).toBe(expected)
+  })
+}

problemset/string-rotation-lcci/index.ts

@@ -0,0 +1,36 @@
+/**
+ * 模拟
+ * @desc 时间复杂度 O(N²)  空间复杂度 O(1)
+ * @param s1
+ * @param s2
+ * @returns
+ */
+export function isFlipedString(s1: string, s2: string): boolean {
+  const m = s1.length
+  const n = s2.length
+  if (m !== n) return false
+  if (n === 0) return true
+
+  for (let i = 0; i < n; i++) {
+    let flag = true
+    for (let j = 0; j < n; j++) {
+      if (s1[(i + j) % n] !== s2[j]) {
+        flag = false
+        break
+      }
+    }
+    if (flag) return true
+  }
+  return false
+}
+
+/**
+ * 搜索子字符串
+ * @desc 时间复杂度 O(N)  空间复杂度 O(N)
+ * @param s1
+ * @param s2
+ * @returns
+ */
+export function isFlipedString2(s1: string, s2: string): boolean {
+  return s1.length === s2.length && (s1 + s1).includes(s2)
+}