feat: leetcode 937

Author: deweyou

README.md

@@ -2,7 +2,7 @@
 
 <p align="center">
 <!-- TOPICS COUNT START -->
-<img src="https://img.shields.io/badge/-进度:352-green" alt="进度:352">
+<img src="https://img.shields.io/badge/-进度:353-green" alt="进度:353">
 <!-- 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

@@ -682,6 +682,12 @@
         "path": "./problemset/count-of-range-sum/README.md",
         "difficulty": "困难"
       },
+      {
+        "id": "937",
+        "title": "重新排列日志文件",
+        "path": "./problemset/reorder-data-in-log-files/README.md",
+        "difficulty": "简单"
+      },
       {
         "id": "954",
         "title": "二倍数对数组",

assets/data/topics.json

@@ -3149,6 +3149,16 @@
     "url": "https://leetcode-cn.com/problems/reverse-only-letters/",
     "path": "./problemset/reverse-only-letters/README.md"
   },
+  {
+    "id": "937",
+    "title": {
+      "cn": "重新排列日志文件",
+      "en": "reorder-data-in-log-files"
+    },
+    "difficulty": "简单",
+    "url": "https://leetcode-cn.com/problems/reorder-data-in-log-files/",
+    "path": "./problemset/reorder-data-in-log-files/README.md"
+  },
   {
     "id": "954",
     "title": {

assets/docs/CATEGORIES.md

@@ -134,6 +134,7 @@
 | 315. [计算右侧小于当前元素的个数](../../problemset/count-of-smaller-numbers-after-self/README.md) | 困难 |
 | 324. [摆动排序 II](../../problemset/wiggle-sort-2/README.md) | 中等 |
 | 327. [区间和的个数](../../problemset/count-of-range-sum/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 |
 | 1984. [学生分数的最小差值](../../problemset/minimum-difference-between-highest-and-lowest-of-k-scores/README.md) | 简单 |
 | 1996. [游戏中弱角色的数量](../../problemset/the-number-of-weak-characters-in-the-game/README.md) | 中等 |

assets/docs/TOPICS.md

@@ -630,6 +630,8 @@
 
 [917. 仅仅反转字母](../../problemset/reverse-only-letters/README.md)
 
+[937. 重新排列日志文件](../../problemset/reorder-data-in-log-files/README.md)
+
 [954. 二倍数对数组](../../problemset/array-of-doubled-pairs/README.md)
 
 [969. 煎饼排序](../../problemset/pancake-sorting/README.md)

problemset/reorder-data-in-log-files/README.md

@@ -0,0 +1,97 @@
+# 重新排列日志文件
+
+> 难度：简单
+>
+> https://leetcode-cn.com/problems/reorder-data-in-log-files/
+
+## 题目
+
+给你一个日志数组 `logs`。每条日志都是以空格分隔的字串，其第一个字为字母与数字混合的 **标识符** 。
+
+有两种不同类型的日志：
+
+- **字母日志**：除标识符之外，所有字均由小写字母组成
+- **数字日志**：除标识符之外，所有字均由数字组成
+
+请按下述规则将日志重新排序：
+
+- 所有 **字母日志** 都排在 **数字日志** 之前。
+- **字母日志** 在内容不同时，忽略标识符后，按内容字母顺序排序；在内容相同时，按标识符排序。
+- **数字日志** 应该保留原来的相对顺序。
+
+返回日志的最终顺序。
+
+### 示例
+
+#### 示例 1：
+
+```
+输入：logs = ["dig1 8 1 5 1","let1 art can","dig2 3 6","let2 own kit dig","let3 art zero"]
+输出：["let1 art can","let3 art zero","let2 own kit dig","dig1 8 1 5 1","dig2 3 6"]
+解释：
+字母日志的内容都不同，所以顺序为 "art can", "art zero", "own kit dig" 。
+数字日志保留原来的相对顺序 "dig1 8 1 5 1", "dig2 3 6" 。
+```
+
+#### 示例 2：
+
+```
+输入：logs = ["a1 9 2 3 1","g1 act car","zo4 4 7","ab1 off key dog","a8 act zoo"]
+输出：["g1 act car","a8 act zoo","ab1 off key dog","a1 9 2 3 1","zo4 4 7"]
+```
+
+## 解题
+
+```ts 
+/**
+ * 自定义排序
+ * @desc 时间复杂度 O(NlogN)  空间复杂度 O(N)
+ * @param logs
+ * @returns
+ */
+export function reorderLogFiles(logs: string[]): string[] {
+  return logs
+    .map<[string, number]>((log, i) => [log, i])
+    .sort((a, b) => logCompare(a, b))
+    .map(item => item[0])
+
+  function logCompare(a: [string, number], b: [string, number]): number {
+    const log1 = split(a[0])
+    const log2 = split(b[0])
+    const isDigit1 = !isNaN(Number(log1[1][0]))
+    const isDigit2 = !isNaN(Number(log2[1][0]))
+    if (isDigit1 && isDigit2)
+      return a[1] - b[1]
+
+    if (!isDigit1 && !isDigit2) {
+      const sc = compareTo(log1[1], log2[1])
+
+      if (sc !== 0) return sc
+      return compareTo(log1[0], log2[0])
+    }
+
+    return isDigit1 ? 1 : -1
+  }
+
+  function split(str: string, separator = ' '): string[] {
+    const arr = str.split(separator)
+    return [arr.shift()!, arr.join(separator)]
+  }
+
+  function compareTo(left: string, right: string): -1 | 0 | 1 {
+    let diff: number
+    for (let i = 0; i < Math.min(left.length, right.length); i++) {
+      diff = left[i].charCodeAt(0) - right[i].charCodeAt(0)
+      if (diff !== 0)
+        return diff < 0 ? -1 : 1
+    }
+
+    diff = left.length - right.length
+    return diff === 0
+      ? 0
+      : diff < 0
+        ? -1
+        : 1
+  }
+}
+```

problemset/reorder-data-in-log-files/index.spec.ts

@@ -0,0 +1,25 @@
+import { describe, expect, it } from 'vitest'
+import { reorderLogFiles } from '.'
+
+describe('重新排列日志文件', () => {
+  testCase(reorderLogFiles)
+})
+
+function testCase(fn: (logs: string[]) => string[]) {
+  it.each([
+    [
+      ['dig1 8 1 5 1', 'let1 art can', 'dig2 3 6', 'let2 own kit dig', 'let3 art zero'],
+      ['let1 art can', 'let3 art zero', 'let2 own kit dig', 'dig1 8 1 5 1', 'dig2 3 6'],
+    ],
+    [
+      ['a1 9 2 3 1', 'g1 act car', 'zo4 4 7', 'ab1 off key dog', 'a8 act zoo'],
+      ['g1 act car', 'a8 act zoo', 'ab1 off key dog', 'a1 9 2 3 1', 'zo4 4 7'],
+    ],
+    [
+      ['1 n u', 'r 527', 'j 893', '6 14', '6 82'],
+      ['1 n u', 'r 527', 'j 893', '6 14', '6 82'],
+    ],
+  ])('示例%#', (logs, expected) => {
+    expect(fn(logs)).toStrictEqual(expected)
+  })
+}

problemset/reorder-data-in-log-files/index.ts

@@ -0,0 +1,51 @@
+/**
+ * 自定义排序
+ * @desc 时间复杂度 O(NlogN)  空间复杂度 O(N)
+ * @param logs
+ * @returns
+ */
+export function reorderLogFiles(logs: string[]): string[] {
+  return logs
+    .map<[string, number]>((log, i) => [log, i])
+    .sort((a, b) => logCompare(a, b))
+    .map(item => item[0])
+
+  function logCompare(a: [string, number], b: [string, number]): number {
+    const log1 = split(a[0])
+    const log2 = split(b[0])
+    const isDigit1 = !isNaN(Number(log1[1][0]))
+    const isDigit2 = !isNaN(Number(log2[1][0]))
+    if (isDigit1 && isDigit2)
+      return a[1] - b[1]
+
+    if (!isDigit1 && !isDigit2) {
+      const sc = compareTo(log1[1], log2[1])
+
+      if (sc !== 0) return sc
+      return compareTo(log1[0], log2[0])
+    }
+
+    return isDigit1 ? 1 : -1
+  }
+
+  function split(str: string, separator = ' '): string[] {
+    const arr = str.split(separator)
+    return [arr.shift()!, arr.join(separator)]
+  }
+
+  function compareTo(left: string, right: string): -1 | 0 | 1 {
+    let diff: number
+    for (let i = 0; i < Math.min(left.length, right.length); i++) {
+      diff = left[i].charCodeAt(0) - right[i].charCodeAt(0)
+      if (diff !== 0)
+        return diff < 0 ? -1 : 1
+    }
+
+    diff = left.length - right.length
+    return diff === 0
+      ? 0
+      : diff < 0
+        ? -1
+        : 1
+  }
+}