feat: leetcode 1805

Author: deweyou

assets/data/categories.json

@@ -2069,6 +2069,12 @@
         "title": "交替合并字符串",
         "path": "./problemset/merge-strings-alternately/README.md",
         "difficulty": "简单"
+      },
+      {
+        "id": "1805",
+        "title": "字符串中不同整数的数目",
+        "path": "./problemset/number-of-different-integers-in-a-string/README.md",
+        "difficulty": "简单"
       }
     ]
   },

assets/data/topics.json

@@ -5279,6 +5279,16 @@
     "url": "https://leetcode.cn/problems/maximum-ascending-subarray-sum/",
     "path": "./problemset/maximum-ascending-subarray-sum/README.md"
   },
+  {
+    "id": "1805",
+    "title": {
+      "cn": "字符串中不同整数的数目",
+      "en": "number-of-different-integers-in-a-string"
+    },
+    "difficulty": "简单",
+    "url": "https://leetcode.cn/problems/number-of-different-integers-in-a-string/",
+    "path": "./problemset/number-of-different-integers-in-a-string/README.md"
+  },
   {
     "id": "1822",
     "title": {

assets/docs/CATEGORIES.md

@@ -382,6 +382,7 @@
 | 1417. [重新格式化字符串](../../problemset/reformat-the-string/README.md) | 简单 |
 | 1455. [检查单词是否为句中其他单词的前缀](../../problemset/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/README.md) | 简单 |
 | 1768. [交替合并字符串](../../problemset/merge-strings-alternately/README.md) | 简单 |
+| 1805. [字符串中不同整数的数目](../../problemset/number-of-different-integers-in-a-string/README.md) | 简单 |
 
 ## 模拟
 

assets/docs/TOPICS.md

@@ -1056,6 +1056,8 @@
 
 [1800. 最大升序子数组和](../../problemset/maximum-ascending-subarray-sum/README.md)
 
+[1805. 字符串中不同整数的数目](../../problemset/number-of-different-integers-in-a-string/README.md)
+
 [1822. 数组元素积的符号](../../problemset/sign-of-the-product-of-an-array/README.md)
 
 [1823. 找出游戏的获胜者](../../problemset/find-the-winner-of-the-circular-game/README.md)

problemset/number-of-different-integers-in-a-string/README.md

@@ -0,0 +1,74 @@
+# 字符串中不同整数的数目
+
+> 难度：简单
+>
+> https://leetcode.cn/problems/number-of-different-integers-in-a-string/
+
+## 题目
+
+给你一个字符串 `word` ，该字符串由数字和小写英文字母组成。
+
+请你用空格替换每个不是数字的字符。例如，"a123bc34d8ef34" 将会变成 " 123  34 8  34" 。注意，剩下的这些整数为（相邻彼此至少有一个空格隔开）："123"、"34"、"8" 和 "34" 。
+
+返回对 `word` 完成替换后形成的 **不同** 整数的数目。
+
+只有当两个整数的 **不含前导零** 的十进制表示不同， 才认为这两个整数也不同。
+
+### 示例
+
+#### 示例 1：
+
+```
+输入：word = "a123bc34d8ef34"
+输出：3
+解释：不同的整数有 "123"、"34" 和 "8" 。注意，"34" 只计数一次。
+```
+
+#### 示例 2：
+
+```
+输入：word = "leet1234code234"
+输出：2
+```
+
+#### 示例 3：
+
+```
+输入：word = "a1b01c001"
+输出：1
+解释："1"、"01" 和 "001" 视为同一个整数的十进制表示，因为在比较十进制值时会忽略前导零的存在。
+```
+
+## 解题
+
+```ts 
+/**
+ * 双指针
+ * @desc 时间复杂度 O(N) 空间复杂度 O(N)
+ * @param word
+ * @returns
+ */
+export function numDifferentIntegers(word: string): number {
+  const isDigit = (ch: string) => ch >= '0' && ch <= '9'
+
+  const set = new Set()
+  const n = word.length; let p1 = 0; let p2
+  while (true) {
+    while (p1 < n && !isDigit(word[p1]))
+      p1++
+    if (p1 === n)
+      break
+
+    p2 = p1
+    while (p2 < n && isDigit(word[p2]))
+      p2++
+
+    while (p2 - p1 > 1 && word[p1] === '0')
+      p1++
+
+    set.add(word.slice(p1, p2))
+    p1 = p2
+  }
+  return set.size
+}
+```

problemset/number-of-different-integers-in-a-string/index.spec.ts

@@ -0,0 +1,16 @@
+import { describe, expect, it } from 'vitest'
+import { numDifferentIntegers } from '.'
+
+describe('字符串中不同整数的数目', () => {
+  testCase(numDifferentIntegers)
+})
+
+function testCase(fn: (word: string) => number) {
+  it.each([
+    ['a123bc34d8ef34', 3],
+    ['leet1234code234', 2],
+    ['a1b01c001', 1],
+  ])('示例%#', (word, expected) => {
+    expect(fn(word)).toBe(expected)
+  })
+}

problemset/number-of-different-integers-in-a-string/index.ts

@@ -0,0 +1,29 @@
+/**
+ * 双指针
+ * @desc 时间复杂度 O(N) 空间复杂度 O(N)
+ * @param word
+ * @returns
+ */
+export function numDifferentIntegers(word: string): number {
+  const isDigit = (ch: string) => ch >= '0' && ch <= '9'
+
+  const set = new Set()
+  const n = word.length; let p1 = 0; let p2
+  while (true) {
+    while (p1 < n && !isDigit(word[p1]))
+      p1++
+    if (p1 === n)
+      break
+
+    p2 = p1
+    while (p2 < n && isDigit(word[p2]))
+      p2++
+
+    while (p2 - p1 > 1 && word[p1] === '0')
+      p1++
+
+    set.add(word.slice(p1, p2))
+    p1 = p2
+  }
+  return set.size
+}