feat: leetcode 556

Author: ouduidui

README.md

@@ -2,7 +2,7 @@
 
 <p align="center">
 <!-- TOPICS COUNT START -->
-<img src="https://img.shields.io/badge/-进度:419-green" alt="进度:419">
+<img src="https://img.shields.io/badge/-进度:420-green" alt="进度:420">
 <!-- 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

@@ -772,6 +772,12 @@
         "path": "./problemset/minimum-moves-to-equal-array-elements-2/README.md",
         "difficulty": "中等"
       },
+      {
+        "id": "556",
+        "title": "下一个更大元素 III",
+        "path": "./problemset/next-greater-element-3/README.md",
+        "difficulty": "中等"
+      },
       {
         "id": "719",
         "title": "找出第 K 小的数对距离",

assets/data/topics.json

@@ -3099,6 +3099,16 @@
     "url": "https://leetcode-cn.com/problems/optimal-division/",
     "path": "./problemset/optimal-division/README.md"
   },
+  {
+    "id": "556",
+    "title": {
+      "cn": "下一个更大元素 III",
+      "en": "next-greater-element-3"
+    },
+    "difficulty": "中等",
+    "url": "https://leetcode.cn/problems/next-greater-element-iii/",
+    "path": "./problemset/next-greater-element-3/README.md"
+  },
   {
     "id": "564",
     "title": {

assets/docs/CATEGORIES.md

@@ -149,6 +149,7 @@
 | 327. [区间和的个数](../../problemset/count-of-range-sum/README.md) | 困难 |
 | 436. [寻找右区间](../../problemset/find-right-interval/README.md) | 中等 |
 | 462. [最少移动次数使数组元素相等 II](../../problemset/minimum-moves-to-equal-array-elements-2/README.md) | 中等 |
+| 556. [下一个更大元素 III](../../problemset/next-greater-element-3/README.md) | 中等 |
 | 719. [找出第 K 小的数对距离](../../problemset/find-k-th-smallest-pair-distance/README.md) | 困难 |
 | 937. [重新排列日志文件](../../problemset/reorder-data-in-log-files/README.md) | 简单 |
 | 954. [二倍数对数组](../../problemset/array-of-doubled-pairs/README.md) | 中等 |

assets/docs/TOPICS.md

@@ -620,6 +620,8 @@
 
 [553. 最优除法](../../problemset/optimal-division/README.md)
 
+[556. 下一个更大元素 III](../../problemset/next-greater-element-3/README.md)
+
 [564. 寻找最近的回文数](../../problemset/find-the-closest-palindrome/README.md)
 
 [587. 安装栅栏](../../problemset/erect-the-fence/README.md)

problemset/next-greater-element-3/README.md

@@ -0,0 +1,64 @@
+# 下一个更大元素 III
+
+> 难度：中等
+>
+> https://leetcode.cn/problems/next-greater-element-iii/
+
+## 题目
+
+给你一个正整数 `n` ，请你找出符合条件的最小整数，其由重新排列 `n` 中存在的每位数字组成，并且其值大于 `n` 。如果不存在这样的正整数，则返回 `-1` 。
+
+注意 ，返回的整数应当是一个 `32` 位整数 ，如果存在满足题意的答案，但不是 `32` 位整数 ，同样返回 `-1` 。
+
+### 示例
+
+#### 示例 1：
+
+```
+输入：n = 12
+输出：21
+```
+
+#### 示例 2：
+```
+输入：n = 21
+输出：-1
+```
+
+## 解题
+
+```ts
+/**
+ * 下一个排列
+ * @desc 时间复杂度 O(logN)  空间复杂度 O(logN)
+ * @param n
+ * @returns
+ */
+export function nextGreaterElement(n: number): number {
+  const MAX = 2 ** 31 - 1
+
+  const nums = [...(`${n}`)]
+  let i = nums.length - 2
+  while (i >= 0 && nums[i] >= nums[i + 1]) i--
+
+  if (i < 0) return -1
+
+  let j = nums.length - 1
+  while (j >= 0 && nums[i] >= nums[j]) j--;
+
+  [nums[i], nums[j]] = [nums[j], nums[i]]
+  reverse(nums, i + 1)
+  const ans = Number(nums.join(''))
+  return ans > MAX ? -1 : ans
+
+  function reverse(nums: string[], begin: number) {
+    let i = begin
+    let j = nums.length - 1
+    while (i < j) {
+      [nums[i], nums[j]] = [nums[j], nums[i]]
+      i++
+      j--
+    }
+  }
+}
+```

problemset/next-greater-element-3/index.spec.ts

@@ -0,0 +1,15 @@
+import { describe, expect, it } from 'vitest'
+import { nextGreaterElement } from '.'
+
+describe('下一个更大元素 III', () => {
+  testCase(nextGreaterElement)
+})
+
+function testCase(fn: (n: number) => number) {
+  it.each([
+    [12, 21],
+    [21, -1],
+  ])('示例%#', (n, expected) => {
+    expect(fn(n)).toBe(expected)
+  })
+}

problemset/next-greater-element-3/index.ts

@@ -0,0 +1,33 @@
+/**
+ * 下一个排列
+ * @desc 时间复杂度 O(logN)  空间复杂度 O(logN)
+ * @param n
+ * @returns
+ */
+export function nextGreaterElement(n: number): number {
+  const MAX = 2 ** 31 - 1
+
+  const nums = [...(`${n}`)]
+  let i = nums.length - 2
+  while (i >= 0 && nums[i] >= nums[i + 1]) i--
+
+  if (i < 0) return -1
+
+  let j = nums.length - 1
+  while (j >= 0 && nums[i] >= nums[j]) j--;
+
+  [nums[i], nums[j]] = [nums[j], nums[i]]
+  reverse(nums, i + 1)
+  const ans = Number(nums.join(''))
+  return ans > MAX ? -1 : ans
+
+  function reverse(nums: string[], begin: number) {
+    let i = begin
+    let j = nums.length - 1
+    while (i < j) {
+      [nums[i], nums[j]] = [nums[j], nums[i]]
+      i++
+      j--
+    }
+  }
+}