feat: 增加 67. 二进制求和

docs/.vitepress/leetcodeItems.ts

@@ -195,6 +195,10 @@ const leetcodeItems = [
     "text": "66. 加一",
     "link": "/leetcode/plusOne"
   },
+  {
+    "text": "67. 二进制求和",
+    "link": "/leetcode/addBinary"
+  },
   {
     "text": "83. 删除排序链表中的重复元素",
     "link": "/leetcode/removeDuplicatesFromSortedList"

docs/leetcode/addBinary.md

@@ -0,0 +1,12 @@
+# 67. 二进制求和
+
+[67. 二进制求和](https://leetcode.cn/problems/add-binary/description/)
+
+## 代码
+
+::: code-group
+
+<<< ../../src/leetcode/addBinary/javascript.js{javascript} [javascript]
+<<< ../../src/leetcode/addBinary/typescript.ts{typescript} [typescript]
+
+:::

docs/leetcode/index.md

@@ -49,6 +49,7 @@
 - [63. 不同路径 II](./uniquePathsIi)
 - [64. 最小路径和](./minimumPathSum)
 - [66. 加一](./plusOne)
+- [67. 二进制求和](./addBinary)
 - [83. 删除排序链表中的重复元素](./removeDuplicatesFromSortedList)
 - [2085. 统计出现过一次的公共字符串](./countCommonWordsWithOneOccurrence)
 - [2182. 构造限制重复的字符串](./constructStringWithRepeatLimit)

src/leetcode/addBinary/javascript.js

@@ -0,0 +1,24 @@
+// 67. 二进制求和：https://leetcode.cn/problems/add-binary/description/
+// 输入：a = "11", b = "1"
+// 输出："100"
+
+export function addBinary (a, b) {
+  const len = Math.max(a.length, b.length)
+  a.length === len ? (b = b.padStart(len, '0')) : (a = a.padStart(len, '0'))
+  let ad = 0
+  let res = ''
+  for (let idx = len - 1; idx >= 0; idx--) {
+    const cur = Number(a[idx]) + Number(b[idx]) + ad
+    if (cur > 1) {
+      res = `${cur % 2}${res}`
+      ad = 1
+    } else {
+      res =  `${cur}${res}`
+      ad = 0
+    }
+  }
+  if (ad > 0) {
+    res =  `${ad}${res}`
+  }
+  return res
+}

src/leetcode/addBinary/typescript.ts

@@ -0,0 +1,24 @@
+// 67. 二进制求和：https://leetcode.cn/problems/add-binary/description/
+// 输入：a = "11", b = "1"
+// 输出："100"
+
+export function addBinary (a: string, b: string): string {
+  const len = Math.max(a.length, b.length)
+  a.length === len ? (b = b.padStart(len, '0')) : (a = a.padStart(len, '0'))
+  let ad = 0
+  let res = ''
+  for (let idx = len - 1; idx >= 0; idx--) {
+    const cur = Number(a[idx]) + Number(b[idx]) + ad
+    if (cur > 1) {
+      res = `${cur % 2}${res}`
+      ad = 1
+    } else {
+      res =  `${cur}${res}`
+      ad = 0
+    }
+  }
+  if (ad > 0) {
+    res =  `${ad}${res}`
+  }
+  return res
+}

test/leetcode/addBinary.test.ts

@@ -0,0 +1,13 @@
+import { expect, test } from 'vitest'
+import { addBinary } from '../../src/leetcode/addBinary/typescript.ts'
+import { addBinary as addBinaryJs } from '../../src/leetcode/addBinary/javascript.js'
+
+test(`addBinary`, () => {
+  expect(addBinary('11', '1')).toBe('100')
+  expect(addBinary('10', '111')).toBe('1001')
+})
+
+test(`addBinaryJs`, () => {
+  expect(addBinaryJs('11', '1')).toBe('100')
+  expect(addBinaryJs('10', '111')).toBe('1001')
+})