feat(rotate-array): 使用额外的数组 | 环形替换 | 数组翻转

Author: oushihao

assets/data/category.json

@@ -1079,6 +1079,12 @@
         "path": "../../problemset/compare-version-numbers/README.md",
         "difficulty": "中等"
       },
+      {
+        "id": 189,
+        "title": "轮转数组",
+        "path": "../../problemset/rotate-array/README.md",
+        "difficulty": "中等"
+      },
       {
         "id": 258,
         "title": "各位相加",

assets/data/problems.json

@@ -1699,6 +1699,16 @@
     "url": "https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-iv/",
     "path": "../../problemset/best-time-to-buy-and-sell-stock-4/README.md"
   },
+  {
+    "id": 189,
+    "title": {
+      "cn": "轮转数组",
+      "en": "rotate-array"
+    },
+    "difficulty": "中等",
+    "url": "https://leetcode-cn.com/problems/rotate-array/",
+    "path": "../../problemset/rotate-array/README.md"
+  },
   {
     "id": 258,
     "title": {

assets/docs/CATEGORY.md

@@ -221,6 +221,7 @@
 | 134. [加油站](../../problemset/gas-station/README.md) | 中等 |
 | 135. [分发糖果](../../problemset/candy/README.md) | 困难 |
 | 165. [比较版本号](../../problemset/compare-version-numbers/README.md) | 中等 |
+| 189. [轮转数组](../../problemset/rotate-array/README.md) | 中等 |
 | 258. [各位相加](../../problemset/add-digits/README.md) | 简单 |
 | 393. [UTF-8 编码验证](../../problemset/utf-8-validation/README.md) | 中等 |
 | 521. [最长特殊序列](../../problemset/longest-uncommon-subsequence/README.md) | 简单 |

assets/docs/PROBLEMS.md

@@ -340,6 +340,8 @@
 
 [188. 买卖股票的最佳时机 IV](../../problemset/best-time-to-buy-and-sell-stock-4/README.md)
 
+[189. 轮转数组](../../problemset/rotate-array/README.md)
+
 [258. 各位相加](../../problemset/add-digits/README.md)
 
 [300. 最长递增子序列](../../problemset/longest-increasing-subsequence/README.md)

problemset/rotate-array/README.md

@@ -0,0 +1,107 @@
+# 轮转数组
+
+> 难度：中等
+>
+> https://leetcode-cn.com/problems/rotate-array/
+
+## 题目
+
+给你一个数组，将数组中的元素向右轮转 `k`  个位置，其中  `k` 是非负数。
+
+### 示例
+
+#### 示例 1:
+
+```
+输入: nums = [1,2,3,4,5,6,7], k = 3
+输出: [5,6,7,1,2,3,4]
+解释:
+向右轮转 1 步: [7,1,2,3,4,5,6]
+向右轮转 2 步: [6,7,1,2,3,4,5]
+向右轮转 3 步: [5,6,7,1,2,3,4]
+```
+
+#### 示例  2:
+
+```
+输入：nums = [-1,-100,3,99], k = 2
+输出：[3,99,-1,-100]
+解释:
+向右轮转 1 步: [99,-1,-100,3]
+向右轮转 2 步: [3,99,-1,-100]
+```
+
+## 解题
+
+### 使用额外的数组
+
+```typescript
+/**
+ * 使用额外的数组
+ * @desc 时间复杂度 O(N) 空间复杂度 O(N)
+ * @param nums
+ * @param k
+ */
+export function rotate(nums: number[], k: number): void {
+  if ((k = k % nums.length) && k === 0) return;
+  nums.unshift(...nums.splice(nums.length - k));
+}
+```
+
+### 环形替换
+
+```typescript
+/**
+ * 环形替换
+ * @desc 时间复杂度 O(N)  空间复杂度 O(1)
+ * @param nums
+ * @param k
+ */
+export function rotate2(nums: number[], k: number): void {
+  const len = nums.length;
+  if ((k = k % len) && k === 0) return;
+
+  // 求最大公约数
+  const gcd = (x: number, y: number): number => (y ? gcd(y, x % y) : x);
+  // 遍历的次数
+  const count = gcd(k, len);
+
+  for (let start = 0; start < count; start++) {
+    let current = start;
+    let prev = nums[start];
+
+    // nums[i] < nums-> nums[(i+k)%len]
+    do {
+      const next = (current + k) % len;
+      [nums[next], prev] = [prev, nums[next]];
+      current = next;
+    } while (start !== current);
+  }
+}
+```
+
+### 数组翻转
+
+```typescript
+/**
+ * 数组翻转
+ * @desc 时间复杂度 O(N)  空间复杂度 O(1)
+ * @param nums
+ * @param k
+ */
+export function rotate3(nums: number[], k: number): void {
+  const len = nums.length;
+  if ((k = k % len) && k === 0) return;
+  reverse(nums, 0, len - 1);
+  reverse(nums, 0, k - 1);
+  reverse(nums, k, len - 1);
+
+  function reverse(nums: number[], start: number, end: number) {
+    while (start < end) {
+      [nums[start], nums[end]] = [nums[end], nums[start]];
+      start++;
+      end--;
+    }
+  }
+}
+```

problemset/rotate-array/index.spec.ts

@@ -0,0 +1,33 @@
+import { rotate, rotate2, rotate3 } from '.';
+
+describe('轮转数组', () => {
+  describe('使用额外的数组', () => {
+    testCase(rotate);
+  });
+
+  describe('环形替换', () => {
+    testCase(rotate2);
+  });
+
+  describe('数组翻转', () => {
+    testCase(rotate3);
+  });
+});
+
+function testCase(fn: (nums: number[], k: number) => void) {
+  it('示例一', () => {
+    const nums = [1, 2, 3, 4, 5, 6, 7];
+    const k = 3;
+    fn(nums, k);
+
+    expect(nums).toStrictEqual([5, 6, 7, 1, 2, 3, 4]);
+  });
+
+  it('示例二', () => {
+    const nums = [-1, -100, 3, 99];
+    const k = 2;
+    fn(nums, k);
+
+    expect(nums).toStrictEqual([3, 99, -1, -100]);
+  });
+}

problemset/rotate-array/index.ts

@@ -0,0 +1,60 @@
+/**
+ * 使用额外的数组
+ * @desc 时间复杂度 O(N) 空间复杂度 O(N)
+ * @param nums
+ * @param k
+ */
+export function rotate(nums: number[], k: number): void {
+  if ((k = k % nums.length) && k === 0) return;
+  nums.unshift(...nums.splice(nums.length - k));
+}
+
+/**
+ * 环形替换
+ * @desc 时间复杂度 O(N)  空间复杂度 O(1)
+ * @param nums
+ * @param k
+ */
+export function rotate2(nums: number[], k: number): void {
+  const len = nums.length;
+  if ((k = k % len) && k === 0) return;
+
+  // 求最大公约数
+  const gcd = (x: number, y: number): number => (y ? gcd(y, x % y) : x);
+  // 遍历的次数
+  const count = gcd(k, len);
+
+  for (let start = 0; start < count; start++) {
+    let current = start;
+    let prev = nums[start];
+
+    // nums[i] < nums-> nums[(i+k)%len]
+    do {
+      const next = (current + k) % len;
+      [nums[next], prev] = [prev, nums[next]];
+      current = next;
+    } while (start !== current);
+  }
+}
+
+/**
+ * 数组翻转
+ * @desc 时间复杂度 O(N)  空间复杂度 O(1)
+ * @param nums
+ * @param k
+ */
+export function rotate3(nums: number[], k: number): void {
+  const len = nums.length;
+  if ((k = k % len) && k === 0) return;
+  reverse(nums, 0, len - 1);
+  reverse(nums, 0, k - 1);
+  reverse(nums, k, len - 1);
+
+  function reverse(nums: number[], start: number, end: number) {
+    while (start < end) {
+      [nums[start], nums[end]] = [nums[end], nums[start]];
+      start++;
+      end--;
+    }
+  }
+}