add: 不同的二叉搜索树

.github/workflows/nodejs.yml

@@ -5,18 +5,17 @@ name: Node.js CI
 
 on:
   push:
-    branches: [ master ]
+    branches: [master]
   pull_request:
-    branches: [ master ]
+    branches: [master]
 
 jobs:
   build:
-
     runs-on: ubuntu-latest
 
     strategy:
       matrix:
-        node-version: [10.x, 12.x]
+        node-version: [12.x, 14.x]
 
     steps:
       - uses: actions/checkout@v2

.vscode/settings.json

@@ -1,21 +1,51 @@
 {
   "cSpell.words": [
+    "ADOBECODEBANC",
+    "Gitpod",
     "LVIII",
+    "Paddr",
+    "Prinme",
     "abcdefg",
+    "abciiidef",
     "cdefgab",
     "chuan",
+    "defang",
+    "defanging",
     "dvdf",
+    "eleetminicoworoep",
+    "inorder",
     "lcci",
     "lcof",
     "lcov",
+    "leetcode",
+    "leetcodeisgreat",
+    "liang",
     "lrloseumgh",
     "mincost",
     "nums",
     "powx",
+    "preorder",
     "pwwkew",
+    "qian",
+    "racecar",
+    "strs",
+    "subarray",
+    "subarrays",
     "umghlrlose",
     "xuan",
+    "zhan",
     "zhong",
     "zhuan"
-  ]
+  ],
+  "standard.options": {
+    "globals": [
+      "$",
+      "jQuery",
+      "fetch"
+    ],
+    "usePackageJson": true,
+    "env": {
+      "jest": true
+    }
+  }
 }

README.md

@@ -63,13 +63,13 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 179. 最大数 <https://leetcode-cn.com/problems/largest-number/>
   - LintCode 184. 最大数 <https://www.lintcode.com/problem/largest-number/description>
 
-- [实现string2int()函数](src/string/string2int.js)
+- [实现 string2int()函数](src/string/string2int.js)
 
-  - 面试题参考思路，不严谨实现 廖雪峰 不要使用JavaScript内置的parseInt()函数，利用map和reduce操作实现一个string2int()函数。 <https://www.liaoxuefeng.com/wiki/1022910821149312/1024322552460832>
+  - 面试题参考思路，不严谨实现 廖雪峰 不要使用 JavaScript 内置的 parseInt()函数，利用 map 和 reduce 操作实现一个 string2int()函数。 <https://www.liaoxuefeng.com/wiki/1022910821149312/1024322552460832>
 
 - [左旋转字符串](src/string/zuo-xuan-zhuan-zi-fu-chuan-lcof.js)
 
-  - LeetCode 面试题58 - II. 左旋转字符串 <https://leetcode-cn.com/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/>
+  - LeetCode 面试题 58 - II. 左旋转字符串 <https://leetcode-cn.com/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/>
 
 - [最后一个单词的长度](src/string/length-of-last-word.js)
 
@@ -86,7 +86,43 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 13. 罗马数字转整数 <https://leetcode-cn.com/problems/roman-to-integer/>
   - LintCode 419. 罗马数字转整数 <https://www.lintcode.com/problem/roman-to-integer/description>
 
-## 数组
+- [验证回文字符串 Ⅱ](src/string/valid-palindrome-ii.js)
+
+  - LeetCode 680. 验证回文字符串 Ⅱ <https://leetcode-cn.com/problems/valid-palindrome-ii/>
+  - LintCode 891. 有效回文 II <https://www.lintcode.com/problem/valid-palindrome-ii/description>
+
+- [IP 地址无效化](src/string/defanging-an-ip-address.js)
+
+  - LeetCode 1108. IP 地址无效化 <https://leetcode-cn.com/problems/defanging-an-ip-address/>
+
+- [最长回文子串](src/string/longest-palindromic-substring.js)
+
+  - LeetCode 5. 最长回文子串 <https://leetcode-cn.com/problems/longest-palindromic-substring/>
+  - LintCode 200. 最长回文子串 <https://www.lintcode.com/problem/longest-palindromic-substring/>
+
+- [每个元音包含偶数次的最长子字符串](src/string/find-the-longest-substring-containing-vowels-in-even-counts.js)
+
+  - LeetCode 1371. 每个元音包含偶数次的最长子字符串 <https://leetcode-cn.com/problems/find-the-longest-substring-containing-vowels-in-even-counts/>
+
+- [最小覆盖子串](src/string/minimum-window-substring.js)
+
+  - LeetCode 76. 最小覆盖子串 <https://leetcode-cn.com/problems/minimum-window-substring/>
+  - LintCode 32. 最小子串覆盖 <https://www.lintcode.com/problem/minimum-window-substring/description>
+
+- [检查单词是否为句中其他单词的前缀](src/string/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence.js)
+
+  - LeetCode 5416. 检查单词是否为句中其他单词的前缀 <https://leetcode-cn.com/problems/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence/>
+
+- [定长子串中元音的最大数目](src/string/maximum-number-of-vowels-in-a-substring-of-given-length.js)
+
+  - LeetCode 5417. 定长子串中元音的最大数目 <https://leetcode-cn.com/problems/maximum-number-of-vowels-in-a-substring-of-given-length/>
+
+- [最长公共前缀](src/string/longest-common-prefix.js)
+
+  - LeetCode 14. 最长公共前缀 <https://leetcode-cn.com/problems/longest-common-prefix/>
+  - LintCode 78. 最长公共前缀 <https://www.lintcode.com/problem/longest-common-prefix/description>
+
+## 数组/队列/集合/映射
 
 - [电话号码的字母组合](src/array/letter-combinations-of-a-phone-number.js)
 
@@ -232,7 +268,7 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 
 - [逆序对](src/array/reverse-pairs.js)
 
-  - LeetCode 面试题51. 数组中的逆序对 <https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/>
+  - LeetCode 面试题 51. 数组中的逆序对 <https://leetcode-cn.com/problems/shu-zu-zhong-de-ni-xu-dui-lcof/>
   - LintCode 532. 逆序对 <https://www.lintcode.com/problem/reverse-pairs/description>
 
 - [搜索旋转排序数组](src/array/search-in-rotated-sorted-array.js)
@@ -242,7 +278,7 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 
 - [数组中数字出现的次数](src/array/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-lcof.js)
 
-  - LeetCode 面试题56 - I. 数组中数字出现的次数 <https://leetcode-cn.com/problems/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-lcof/>
+  - LeetCode 面试题 56 - I. 数组中数字出现的次数 <https://leetcode-cn.com/problems/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-lcof/>
 
 - [只出现一次的数字](src/array/single-number.js)
 
@@ -262,10 +298,10 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 - [最大子序和](src/array/maximum-subarray.js)
 
   - LeetCode 53. 最大子序和 <https://leetcode-cn.com/problems/maximum-subarray/>
-  - LeetCode 面试题42. 连续子数组的最大和 <https://leetcode-cn.com/problems/lian-xu-zi-shu-zu-de-zui-da-he-lcof/>
+  - LeetCode 面试题 42. 连续子数组的最大和 <https://leetcode-cn.com/problems/lian-xu-zi-shu-zu-de-zui-da-he-lcof/>
   - LintCode 41. 最大子数组 <https://www.lintcode.com/problem/maximum-subarray/description>
 
-- [跳跃游戏II](src/array/jump-game-ii.js)
+- [跳跃游戏 II](src/array/jump-game-ii.js)
 
   - LeetCode 45. 跳跃游戏 II <https://leetcode-cn.com/problems/jump-game-ii/>
   - LintCode 117. 跳跃游戏 II <https://www.lintcode.com/problem/jump-game-ii/description>
@@ -283,6 +319,73 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 221. 最大正方形 <https://leetcode-cn.com/problems/maximal-square/>
   - LintCode 436. 最大正方形 <https://www.lintcode.com/problem/maximal-square/description>
 
+- [数据流中第一个唯一的数字](src/array/first-unique-number-in-data-stream.js)
+
+  - LintCode 685. 数据流中第一个唯一的数字 <https://www.lintcode.com/problem/first-unique-number-in-data-stream/description>
+
+- [第一个只出现一次的字符](src/array/first-unique-character-in-a-string.js)
+
+  - LeetCode 面试题 50. 第一个只出现一次的字符 <https://leetcode-cn.com/problems/di-yi-ge-zhi-chu-xian-yi-ci-de-zi-fu-lcof/>
+  - LeetCode 387. 字符串中的第一个唯一字符 <https://leetcode-cn.com/problems/first-unique-character-in-a-string/>
+  - LintCode 209. 第一个只出现一次的字符 <https://www.lintcode.com/problem/first-unique-character-in-a-string/description>
+
+- [乘积最大子数组](src/array/maximum-product-subarray.js)
+
+  - LeetCode 152. 乘积最大子数组 <https://leetcode-cn.com/problems/maximum-product-subarray/>
+  - LintCode 191. 乘积最大子序列 <https://www.lintcode.com/problem/maximum-product-subarray/description>
+
+- [最高频率的 IP](src/array/highest-frequency-ip.js)
+
+  - LintCode 1613. 最高频率的 IP <https://www.lintcode.com/problem/highest-frequency-ip/description>
+
+- [柠檬水找零](src/array/lemonade-change.js)
+
+  - LeetCode 860. 柠檬水找零 <https://leetcode-cn.com/problems/lemonade-change/>
+  - LintCode 1509. 柠檬水找零 <https://www.lintcode.com/problem/lemonade-change/description>
+
+- [LRU 缓存机制](src/array/lru-cache.js)
+
+  - LeetCode 146. LRU 缓存机制 <https://leetcode-cn.com/problems/lru-cache/>
+  - LintCode 134. LRU 缓存策略 <https://www.lintcode.com/problem/lru-cache/description>
+
+- [寻找重复数](src/array/find-the-duplicate-number.js)
+
+  - LeetCode 287. 寻找重复数 <https://leetcode-cn.com/problems/find-the-duplicate-number/>
+  - LintCode 633. 寻找重复的数 <https://www.lintcode.com/problem/find-the-duplicate-number/description>
+
+- [和可被 K 整除的子数组](src/array/subarray-sums-divisible-by-k.js)
+
+  - LeetCode 974. 和可被 K 整除的子数组 <https://leetcode-cn.com/problems/subarray-sums-divisible-by-k/>
+
+- [打家劫舍](src/array/house-robber.js)
+
+  - LeetCode 198. 打家劫舍 <https://leetcode-cn.com/problems/house-robber/>
+  - LintCode 392. 打劫房屋 <https://www.lintcode.com/problem/house-robber/description>
+
+- [拥有最多糖果的孩子](src/array/kids-with-the-greatest-number-of-candies.js)
+
+  - LeetCode 1431. 拥有最多糖果的孩子 <https://leetcode-cn.com/problems/kids-with-the-greatest-number-of-candies/>
+
+- [除自身以外数组的乘积](src/array/product-of-array-except-self.js)
+
+  - LeetCode 238. 除自身以外数组的乘积 <https://leetcode-cn.com/problems/product-of-array-except-self/>
+  - LintCode 1310. 数组除了自身的乘积 <https://www.lintcode.com/problem/product-of-array-except-self/description>
+
+- [每日温度](src/array/daily-temperatures.js)
+
+  - LeetCode 739. 每日温度 <https://leetcode-cn.com/problems/daily-temperatures/>
+  - LintCode 1060. 每日温度 <https://www.lintcode.com/problem/daily-temperatures/description>
+
+- [最长重复子数组](src/array/maximum-length-of-repeated-subarray.js)
+
+  - LeetCode 718. 最长重复子数组 <https://leetcode-cn.com/problems/maximum-length-of-repeated-subarray/>
+  - LintCode 79. 最长公共子串 <https://www.lintcode.com/problem/longest-common-substring/description>
+
+- [有序矩阵中第 K 小的元素](src/array/kth-smallest-element-in-a-sorted-matrix.js)
+
+  - LeetCode 378. 有序矩阵中第 K 小的元素 <https://leetcode-cn.com/problems/kth-smallest-element-in-a-sorted-matrix/>
+  - LintCode 401. 排序矩阵中的从小到大第 k 个数 <https://www.lintcode.com/problem/kth-smallest-number-in-sorted-matrix/description>
+
 ## 栈
 
 - [最大矩阵](src/stack/maximal-rectangle.js)
@@ -295,6 +398,15 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 155. 最小栈 <https://leetcode-cn.com/problems/min-stack/>
   - LintCode 12. 带最小值操作的栈 <https://www.lintcode.com/problem/min-stack/description>
 
+- [用两个栈实现队列](src/stack/yong-liang-ge-zhan-shi-xian-dui-lie-lcof.js)
+
+  - LeetCode 剑指 Offer 09. 用两个栈实现队列 <https://leetcode-cn.com/problems/yong-liang-ge-zhan-shi-xian-dui-lie-lcof/>
+
+- [字符串解码](src/stack/decode-string.js)
+
+  - LeetCode 394. 字符串解码 <https://leetcode-cn.com/problems/decode-string/>
+  - LintCode 575. 字符串解码 <https://www.lintcode.com/problem/decode-string/description>
+
 ## 数学
 
 - [阶乘后的零](src/math/factorial-trailing-zeroes.js)
@@ -323,12 +435,37 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 - [回文数](src/math/palindrome-number.js)
 
   - LeetCode 9. 回文数 <https://leetcode-cn.com/problems/palindrome-number/>
-  - LintCode 491. 回文数 <https://www.lintcode.com/problem/palindrome-number/>
+  - LintCode 491. 回文数 <https://www.lintcode.com/problem/palindrome-number/description>
 
 - [Pow(x, n)](src/math/powx-n.js)
 
   - LeetCode 50. Pow(x, n) <https://leetcode-cn.com/problems/powx-n/>
-  - LintCode 428. x的n次幂 <https://www.lintcode.com/problem/powx-n/>
+  - LintCode 428. x 的 n 次幂 <https://www.lintcode.com/problem/powx-n/description>
+
+- [和为 K 的子数组](src/math/subarray-sum-equals-k.js)
+
+  - LeetCode 560. 和为 K 的子数组 <https://leetcode-cn.com/problems/subarray-sum-equals-k/>
+  - LintCode 838. 子数组和为 K <https://www.lintcode.com/problem/subarray-sum-equals-k/description>
+
+- [完美数](src/math/perfect-number.js)
+
+  - LeetCode 507. 完美数 <https://leetcode-cn.com/problems/perfect-number/>
+  - LintCode 1199. 完美的数 <https://www.lintcode.com/problem/perfect-number/description>
+
+- [寻找两个正序数组的中位数](src/math/median-of-two-sorted-arrays.js)
+
+  - LeetCode 4. 寻找两个正序数组的中位数 <https://leetcode-cn.com/problems/median-of-two-sorted-arrays/>
+  - LintCode 65. 两个排序数组的中位数 <https://www.lintcode.com/problem/median-of-two-sorted-arrays/description>
+
+- [三数之和](src/math/3sum.js)
+
+  - LeetCode 15. 三数之和 <https://leetcode-cn.com/problems/3sum/>
+  - LintCode 57. 三数之和 <https://www.lintcode.com/problem/3sum/description>
+
+- [二进制求和](src/math/add-binary.js)
+
+  - LeetCode 67. 二进制求和 <https://leetcode-cn.com/problems/add-binary/>
+  - LintCode 408. 二进制求和 <https://www.lintcode.com/problem/add-binary/description>
 
 ## 堆
 
@@ -347,7 +484,7 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 - [验证二叉搜索树](src/tree/validate-binary-search-tree.js)
 
   - LeetCode 98. 验证二叉搜索树 <https://leetcode-cn.com/problems/validate-binary-search-tree/>
-  - LintCode 95. 验证二叉查找树 <https://www.lintcode.com/problem/validate-binary-search-tree/>
+  - LintCode 95. 验证二叉查找树 <https://www.lintcode.com/problem/validate-binary-search-tree/description>
 
 - [二叉树的最近公共祖先](src/tree/lowest-common-ancestor-of-a-binary-tree.js)
 
@@ -359,17 +496,41 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
   - LeetCode 572. 另一个树的子树 <https://leetcode-cn.com/problems/subtree-of-another-tree/>
   - LintCode 1165. 另一个树的子树 <https://www.lintcode.com/problem/subtree-of-another-tree/description>
 
+- [二叉树的层序遍历](src/tree/binary-tree-level-order-traversal.js)
+
+  - LeetCode 102. 二叉树的层序遍历 <https://leetcode-cn.com/problems/binary-tree-level-order-traversal/>
+  - LintCode 69. 二叉树的层次遍历 <https://www.lintcode.com/problem/binary-tree-level-order-traversal/description>
+
+- [二叉树的层次遍历 II](src/tree/binary-tree-level-order-traversal-ii.js)
+
+  - LeetCode 107. 二叉树的层次遍历 II <https://leetcode-cn.com/problems/binary-tree-level-order-traversal-ii/>
+  - LintCode 70. 二叉树的层次遍历 II <https://www.lintcode.com/problem/binary-tree-level-order-traversal-ii/description>
+
+- [从前序与中序遍历序列构造二叉树](src/tree/construct-binary-tree-from-preorder-and-inorder-traversal.js)
+
+  - LeetCode 105. 从前序与中序遍历序列构造二叉树 <https://leetcode-cn.com/problems/construct-binary-tree-from-preorder-and-inorder-traversal/>
+  - LintCode 73. 前序遍历和中序遍历树构造二叉树 <https://www.lintcode.com/problem/construct-binary-tree-from-preorder-and-inorder-traversal/description>
+
+- [将有序数组转换为二叉搜索树](src/tree/convert-sorted-array-to-binary-search-tree.js)
+
+  - LeetCode 108. 将有序数组转换为二叉搜索树 <https://leetcode-cn.com/problems/convert-sorted-array-to-binary-search-tree/>
+  - LintCode 106. 有序链表转换为二叉搜索树 <https://www.lintcode.com/problem/convert-sorted-list-to-binary-search-tree/description>
+
+- [不同的二叉搜索树](src/tree/unique-binary-search-trees.js)
+
+  - LeetCode 96. 不同的二叉搜索树 <https://leetcode-cn.com/problems/unique-binary-search-trees/>
+
 ## 链表
 
-- [合并K个排序链表](src/list/merge-k-sorted-lists.js)
+- [合并 K 个排序链表](src/list/merge-k-sorted-lists.js)
 
-  - LeetCode 23. 合并K个排序链表 <https://leetcode-cn.com/problems/merge-k-sorted-lists/>
-  - LintCode 104. 合并k个排序链表 <https://www.lintcode.com/problem/merge-k-sorted-lists/description>
+  - LeetCode 23. 合并 K 个排序链表 <https://leetcode-cn.com/problems/merge-k-sorted-lists/>
+  - LintCode 104. 合并 k 个排序链表 <https://www.lintcode.com/problem/merge-k-sorted-lists/description>
 
 - [合并两个有序链表](src/list/merge-two-sorted-lists.js)
 
   - LeetCode 21. 合并两个有序链表 <https://leetcode-cn.com/problems/merge-two-sorted-lists/>
-  - LeetCode 面试题25. 合并两个排序的链表 <https://leetcode-cn.com/problems/he-bing-liang-ge-pai-xu-de-lian-biao-lcof>
+  - LeetCode 面试题 25. 合并两个排序的链表 <https://leetcode-cn.com/problems/he-bing-liang-ge-pai-xu-de-lian-biao-lcof>
   - LintCode 165. 合并两个排序链表 <https://www.lintcode.com/problem/merge-two-sorted-lists/description>
 
 - [链表排序](src/list/sort-list.js)
@@ -386,3 +547,16 @@ LeetCode 与 LintCode 解题记录。此为个人练习仓库，代码中对重
 
   - LeetCode 876. 链表的中间结点 <https://leetcode-cn.com/problems/middle-of-the-linked-list/>
   - LintCode 1609. 链表的中间结点 <https://www.lintcode.com/problem/middle-of-the-linked-list/description>
+
+- [K 个一组翻转链表](src/list/reverse-nodes-in-k-group.js)
+
+  - LeetCode 25. K 个一组翻转链表 <https://leetcode-cn.com/problems/reverse-nodes-in-k-group/>
+  - LintCode 450. K 组翻转链表 <https://www.lintcode.com/problem/reverse-nodes-in-k-group/description>
+
+## 图
+
+- [课程表](src/graphs/course-schedule-ii.js)
+
+  - LeetCode 210. 课程表 II <https://leetcode-cn.com/problems/course-schedule-ii/>
+  - LeetCode 207. 课程表 <https://leetcode-cn.com/problems/course-schedule/>
+  - LintCode 615. 课程表 <https://www.lintcode.com/problem/course-schedule/description>

src/array/daily-temperatures.js

@@ -0,0 +1,18 @@
+/**
+ * @param {number[]} T
+ * @return {number[]}
+ */
+export const dailyTemperatures = function (T) {
+  const len = T.length
+  const ans = new Array(len).fill(0)
+  const stack = []
+  for (let i = 0; i < len; i++) {
+    const temperature = T[i]
+    while (stack.length && temperature > T[stack[stack.length - 1]]) {
+      const prevIndex = stack.pop()
+      ans[prevIndex] = i - prevIndex
+    }
+    stack.push(i)
+  }
+  return ans
+}

src/array/find-the-duplicate-number.js

@@ -0,0 +1,24 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+export const findDuplicate = function (nums) {
+  const len = nums.length
+  let l = 0; let r = len - 1; let res = -1 // 数字都在1-n
+  while (l <= r) {
+    const mid = (l + r) >> 1
+    let cnt = 0
+    for (let i = 0; i < len; i++) {
+      cnt += nums[i] <= mid // true = 1，满足条件才计数
+    }
+
+    if (cnt <= mid) {
+      l = mid + 1
+    } else {
+      r = mid - 1
+      res = mid
+    }
+  }
+
+  return res
+}

src/array/first-unique-character-in-a-string.js

@@ -0,0 +1,18 @@
+/**
+ * @param {string} s
+ * @return {character}
+ */
+export const firstUniqChar = function (s) {
+  const due = new Set()
+  const queue = new Set()
+  for (const n of s) {
+    if (queue.has(n)) {
+      queue.delete(n)
+      due.add(n)
+    } else if (!due.has(n)) {
+      queue.add(n)
+    }
+  }
+
+  return Array.from(queue)[0] || ' '
+}

src/array/first-unique-number-in-data-stream.js

@@ -0,0 +1,23 @@
+/**
+ * @param nums: a continuous stream of numbers
+ * @param number: a number
+ * @return: returns the first unique number
+ */
+export const firstUniqueNumber = function (nums, number) {
+  const due = new Set()
+  const queue = new Set()
+  for (const n of nums) {
+    if (queue.has(n)) {
+      queue.delete(n)
+      due.add(n)
+    } else if (!due.has(n)) {
+      queue.add(n)
+    }
+
+    if (n === number) break
+  }
+
+  if (!due.has(number) && !queue.has(number)) return -1
+
+  return Array.from(queue)[0] || -1
+}

src/array/highest-frequency-ip.js

@@ -0,0 +1,19 @@
+/**
+ * @param ipLines: ip  address
+ * @return: return highestFrequency ip address
+ */
+export const highestFrequency = function (ipLines) {
+  const map = new Map()
+  let max = 0
+  let maxN = ipLines[0]
+  ipLines.forEach(n => {
+    const temp = (map.get(n) || 0) + 1
+    map.set(n, temp)
+    if (temp > max) {
+      maxN = n
+      max = temp
+    }
+  })
+
+  return maxN
+}

src/array/house-robber.js

@@ -0,0 +1,18 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+export const rob = function (nums) {
+  if (!nums) return 0
+  const len = nums.length
+  if (len === 0) return 0
+  if (len === 1) return nums[0]
+
+  let first = nums[0]; let second = Math.max(nums[0], nums[1])
+  for (let i = 2; i < len; i++) {
+    const temp = second
+    second = Math.max(first + nums[i], second)
+    first = temp
+  }
+  return second
+}

src/array/kids-with-the-greatest-number-of-candies.js

@@ -0,0 +1,8 @@
+/**
+ * @param {number[]} candies
+ * @param {number} extraCandies
+ * @return {boolean[]}
+ */
+export const kidsWithCandies = (candies, extraCandies) => {
+  return candies.map(n => n + extraCandies >= Math.max(...candies))
+}

src/array/kth-smallest-element-in-a-sorted-matrix.js

@@ -0,0 +1,8 @@
+/**
+ * @param {number[][]} matrix
+ * @param {number} k
+ * @return {number}
+ */
+export const kthSmallest = function (matrix, k) {
+  return matrix.flat().sort((a, b) => a - b)[k - 1]
+}

src/array/lemonade-change.js

@@ -0,0 +1,31 @@
+/**
+ * @param {number[]} bills
+ * @return {boolean}
+ */
+export const lemonadeChange = function (bills) {
+  const map = new Map()
+
+  while (bills.length) {
+    const money = bills.shift()
+
+    if (money === 5) {
+      map.set(5, (map.get(5) || 0) + 1)
+    } else {
+      let change = money - 5
+      while (change !== 0 && change - 10 > 0 && map.get(10) > 0) {
+        map.set(10, map.get(10) - 1)
+        change -= 10
+      }
+
+      while (change !== 0 && map.get(5) > 0) {
+        map.set(5, map.get(5) - 1)
+        change -= 5
+      }
+
+      if (change !== 0) return false
+      map.set(money, (map.get(money) || 0) + 1)
+    }
+  }
+
+  return true
+}

src/array/lru-cache.js

@@ -0,0 +1,21 @@
+export default class LRUCache {
+  constructor (capacity) {
+    this.max = capacity; this.map = new Map()
+  }
+
+  get (key) {
+    const value = this.map.get(key)
+    if (value !== undefined) {
+      this.map.delete(key)
+      this.map.set(key, value)
+      return value
+    }
+    return -1
+  }
+
+  put (key, value) {
+    this.map.delete(key)
+    this.map.set(key, value)
+    if (this.map.size > this.max) this.map.delete(this.map.keys().next().value)
+  }
+}

src/array/maximum-length-of-repeated-subarray.js

@@ -0,0 +1,38 @@
+// 输入:
+// A: [1, 2, 3, 2, 1]
+// B: [3, 2, 1, 4, 7]
+
+const maxLength = function (A, B, addA, addB, len) {
+  addA = (addA > 0) ? addA : 0
+  addB = (addB > 0) ? addB : 0
+  let result = 0
+  let k = 0
+  for (let i = 0; i < len && (k + len - i > result); i++) {
+    if (A[i + addA] === B[i + addB]) {
+      k++
+    } else {
+      k = 0
+    }
+    result = Math.max(result, k)
+  }
+  return result
+}
+
+/**
+ * @param {number[]} A
+ * @param {number[]} B
+ * @return {number}
+ */
+export const findLength = function (A, B) {
+  const ALen = A.length
+  const BLen = B.length
+  let result = 0
+  for (let i = 1; i < ALen + BLen; i++) {
+    if (result >= (ALen + BLen - i)) {
+      return result
+    }
+    const len = Math.min(i, ALen, BLen, (ALen + BLen - i))
+    result = Math.max(maxLength(A, B, ALen - i, i - ALen, len), result)
+  }
+  return result
+}

src/array/maximum-product-subarray.js

@@ -0,0 +1,18 @@
+/**
+ * @param {number[]} nums
+ * @return {number}
+ */
+export const maxProduct = function (nums) {
+  let res = nums[0]
+  let prevMin = nums[0]
+  let prevMax = nums[0]
+  let tmp1 = 0; let tmp2 = 0
+  for (let i = 1; i < nums.length; i++) {
+    tmp1 = prevMin * nums[i]
+    tmp2 = prevMax * nums[i]
+    prevMin = Math.min(tmp1, tmp2, nums[i])
+    prevMax = Math.max(tmp1, tmp2, nums[i])
+    res = Math.max(prevMax, res)
+  }
+  return res
+}

src/array/product-of-array-except-self.js

@@ -0,0 +1,23 @@
+/**
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+export const productExceptSelf = function (nums) {
+  const ans = []
+  for (let i = 0, len = nums.length; i < len; i++) {
+    let tmpL = 1
+    let tmpR = 1
+    for (let j = 0; j < len; j++) {
+      if (j < i) {
+        tmpL *= nums[j]
+      } else if (j > i) {
+        tmpR *= nums[j]
+      }
+    }
+    ans.push(tmpL * tmpR)
+  }
+
+  return ans
+}
+
+// 此算法效率不高，请勿仿照作者偷懒

src/array/subarray-sums-divisible-by-k.js

@@ -0,0 +1,17 @@
+/**
+ * @param {number[]} A
+ * @param {number} K
+ * @return {number}
+ */
+export const subarraysDivByK = function (A, K) {
+  const map = new Map([[0, 1]]); let sum = 0; let count = 0
+
+  A.forEach(n => {
+    sum = (sum + n) % K; sum = sum < 0 ? sum + K : sum
+    const temp = map.get(sum) || 0
+    count += temp
+    map.set(sum, temp + 1)
+  })
+
+  return count
+}

src/graphs/course-schedule-ii.js

@@ -0,0 +1,34 @@
+/**
+ * @param {number} numCourses
+ * @param {number[][]} prerequisites
+ * @return {number[]}
+ */
+export const findOrder = function (numCourses, prerequisites) {
+  const courses = Array(numCourses).fill(0) // 初始化 上课 需要先完成课程 的门数
+  const obj = {} // 记录受该课程 影响的其他课
+  prerequisites.forEach(item => {
+    const one = item[0]; const two = item[1] // one 要上的课, two 需先完成的课
+    courses[one]++ // 门数 + 1
+    obj[two] ? obj[two].push(one) : obj[two] = [one] // 存在就加, 不存在就新建
+  })
+  const res = []
+  const queue = [] // 队列
+  courses.forEach((t, i) => { // 往队列添加 无需先上 就可以 上 的课
+    if (t === 0) queue.push(i) // 因为是从0开始的, 所以索引也能代替 课的名称
+  })
+  while (queue.length) {
+    const cur = queue.shift() // 出队   表示该课已经上了
+    res.push(cur) // 把出队的放入 结果数组
+    const list = obj[cur] // 获取受该课影响的 课
+    list && list.forEach(item => {
+      courses[item]-- // 因为 出队表示该课已经上了, 所以 要先完成的门数 - 1
+      if (courses[item] === 0) { // 当这个课 要先修完的 已经修完了, 入队
+        queue.push(item)
+      }
+    })
+  }
+
+  return res.length === numCourses ? res : []
+}
+
+// 参考自 https://leetcode-cn.com/problems/course-schedule-ii/solution/chao-da-an-by-shetia/

src/list/reverse-nodes-in-k-group.js

@@ -0,0 +1,48 @@
+/**
+ * Definition for singly-linked list.
+ * function ListNode(val) {
+ *     this.val = val;
+ *     this.next = null;
+ * }
+ */
+/**
+ * @param {ListNode} head
+ * @param {number} k
+ * @return {ListNode}
+ */
+export const reverseKGroup = function (head, k) {
+  // 《K个一组翻转链表》 https://www.wyr.me/post/622
+  let sum = 0 // 记录进行的结点个数
+  let start = head // 记录每次翻转的第一个元素
+  let res = head // 返回值：如果进行过翻转，则为第一次翻转的最后一个结点
+  const queue = [] // 使用队列，方便连接上一次翻转的链表，最大长度为2
+  while (head) { // 遍历一次链表
+    if (++sum === k) { // 如果经过了k个结点，则翻转从start到head的一段结点
+      const headNext = head.next
+      queue.push(start) // 计入队列
+      let next = head.next
+      for (let i = 0; i < sum - 1; i++) { // 翻转结点
+        const tmp = start.next
+        start.next = next
+        next = start
+        start = tmp
+      }
+
+      start.next = next // 最后一个结点
+
+      if (queue.length === 1) { // 判断是否为第一次翻转
+        res = start
+      } else {
+        const la = queue.shift() // 连接上一次翻转的链表
+        la.next = head
+      }
+      sum = 0 // 重置计数
+      start = headNext
+      head = headNext
+    } else {
+      head = head.next
+    }
+  }
+
+  return res
+}

src/math/3sum.js

@@ -0,0 +1,31 @@
+// 一年前做过。复制自：https://leetcode-cn.com/problems/3sum/solution/three-sum-ti-jie-by-wonderful611/
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+export const threeSum = function (nums) {
+  const res = []
+  const length = nums.length
+  nums.sort((a, b) => a - b) // 先排个队，最左边是最弱（小）的，最右边是最强(大)的
+  if (nums[0] <= 0 && nums[length - 1] >= 0) { // 优化1: 整个数组同符号，则无解
+    for (let i = 0; i < length - 2;) {
+      if (nums[i] > 0) break // 优化2: 最左值为正数则一定无解
+      let first = i + 1
+      let last = length - 1
+      do {
+        if (first >= last || nums[i] * nums[last] > 0) break // 两人选相遇，或者三人同符号，则退出
+        const result = nums[i] + nums[first] + nums[last]
+        if (result === 0) { // 如果可以组队
+          res.push([nums[i], nums[first], nums[last]])
+        }
+        if (result <= 0) { // 实力太弱，把菜鸟那边右移一位
+          while (first < last && nums[first] === nums[++first]) { } // 如果相等就跳过
+        } else { // 实力太强，把大神那边右移一位
+          while (first < last && nums[last] === nums[--last]) { }
+        }
+      } while (first < last)
+      while (nums[i] === nums[++i]) { }
+    }
+  }
+  return res
+}

src/math/add-binary.js

@@ -0,0 +1,9 @@
+/**
+ * @param {string} a
+ * @param {string} b
+ * @return {string}
+ */
+export const addBinary = function (a, b) {
+  // eslint-disable-next-line no-undef
+  return (BigInt('0b' + a) + BigInt('0b' + b)).toString(2)
+}

src/math/median-of-two-sorted-arrays.js

@@ -0,0 +1,52 @@
+// 中位数：将一个集合划分为两个长度相等的子集，其中一个子集中的元素总是大于另一个子集中的元素。
+// 参考：https://leetcode-cn.com/problems/median-of-two-sorted-arrays/solution/xun-zhao-liang-ge-you-xu-shu-zu-de-zhong-wei-s-114/
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number}
+ */
+export const findMedianSortedArrays = function (nums1, nums2) {
+  if (nums1.length > nums2.length) return findMedianSortedArrays(nums2, nums1)
+
+  const m = nums1.length; const n = nums2.length
+  let left = 0; let right = m
+  let median1 = 0; let median2 = 0
+
+  while (left <= right) {
+    const i = (left + right) >> 1
+    const j = ((m + n + 1) >> 1) - i
+
+    const numsIm1 = (i === 0 ? Number.MIN_SAFE_INTEGER : nums1[i - 1]) // nums1[i-1]
+    const numsI = (i === m ? Number.MAX_SAFE_INTEGER : nums1[i]) // nums1[i]
+    const numsJm1 = (j === 0 ? Number.MIN_SAFE_INTEGER : nums2[j - 1]) // nums2[j-1]
+    const numsJ = (j === n ? Number.MAX_SAFE_INTEGER : nums2[j]) // nums2[j]
+
+    if (numsIm1 <= numsJ) {
+      median1 = Math.max(numsIm1, numsJm1)
+      median2 = Math.min(numsI, numsJ)
+      left = i + 1
+    } else {
+      right = i - 1
+    }
+  }
+
+  return (m + n) % 2 === 0 ? (median1 + median2) / 2.0 : median1
+}
+
+// 常规解法，不满足复杂度
+// /**
+//  * @param {number[]} nums1
+//  * @param {number[]} nums2
+//  * @return {number}
+//  */
+// var findMedianSortedArrays = function(nums1, nums2) {
+//   const newArr = nums1.concat(nums2).sort((a, b) => a - b)
+//   const length = newArr.length
+
+//   if (length % 2 === 0) {
+//     const half = length / 2
+//     return (newArr[half - 1] + newArr[half]) / 2
+//   } else {
+//     return newArr[(length - 1) / 2]
+//   }
+// };

src/math/perfect-number.js

@@ -0,0 +1,16 @@
+const pn = (p) => { // 欧几里得-欧拉定理
+  return (1 << (p - 1)) * ((1 << p) - 1)
+}
+
+/**
+ * @param {number} num
+ * @return {boolean}
+ */
+export const checkPerfectNumber = function (num) {
+  const primes = [2, 3, 5, 7, 13, 17, 19, 31]
+  for (const prime of primes) {
+    if (pn(prime) === num) return true
+  }
+
+  return false
+}

src/math/subarray-sum-equals-k.js

@@ -0,0 +1,25 @@
+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+export const subarraySum = function (nums, k) {
+  const map = new Map()
+  let sum = 0
+  let res = 0
+
+  nums.forEach((n, index) => {
+    sum += n
+    if (sum === k) res++
+    const subSum = sum - k
+    let value = []
+    if (map.has(subSum)) res += map.get(subSum).length
+    if (map.has(sum)) {
+      value = map.get(sum)
+    }
+    value.push(index)
+    map.set(sum, value)
+  })
+
+  return res
+}

src/stack/decode-string.js

@@ -0,0 +1,36 @@
+// s = "3[a]2[bc]", 返回 "aaabcbc".
+// s = "3[a2[c]]", 返回 "accaccacc".
+// s = "2[abc]3[cd]ef", 返回 "abcabccdcdcdef".
+/**
+ * @param {string} s
+ * @return {string}
+ */
+export const decodeString = function (s) {
+  const stack = []
+  let multiple = ''
+  for (let i = 0, len = s.length; i < len; i++) {
+    if (!isNaN(Number(s[i]))) { // 判断是数字
+      if (i === 0 || !isNaN(Number(s[i - 1]))) { // 如果上一位也是数字，则
+        multiple += s[i]
+      } else {
+        multiple = s[i]
+      }
+    } else if (multiple && s[i] === '[') {
+      stack.push(Number(multiple))
+      multiple = ''
+    } else if (s[i] === ']') {
+      let current = stack.pop()
+      let tmpStr = ''
+      while (typeof current !== 'number') {
+        tmpStr = current + tmpStr
+        current = stack.pop()
+      }
+      tmpStr = tmpStr.repeat(current)
+      stack.push(tmpStr)
+    } else {
+      stack.push(s[i])
+    }
+  }
+
+  return stack.join('')
+}

src/stack/yong-liang-ge-zhan-shi-xian-dui-lie-lcof.js

@@ -0,0 +1,37 @@
+export var CQueue = function () {
+  this.stackA = []
+  this.stackB = []
+}
+
+/**
+ * @param {number} value
+ * @return {void}
+ */
+CQueue.prototype.appendTail = function (value) {
+  this.stackA.push(value)
+}
+
+/**
+ * @return {number}
+ */
+CQueue.prototype.deleteHead = function () {
+  const stackBNum = this.stackB.length
+
+  while (this.stackA.length) {
+    this.stackB.push(this.stackA.pop())
+  }
+
+  if (stackBNum !== 0) {
+    this.stackB.push(...this.stackB.splice(0, stackBNum))
+  }
+
+  const tmp = this.stackB.pop()
+  return tmp === undefined ? -1 : tmp
+}
+
+/**
+ * Your CQueue object will be instantiated and called as such:
+ * var obj = new CQueue()
+ * obj.appendTail(value)
+ * var param_2 = obj.deleteHead()
+ */

src/string/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence.js

@@ -0,0 +1,14 @@
+/**
+ * @param {string} sentence
+ * @param {string} searchWord
+ * @return {number}
+ */
+export const isPrefixOfWord = function (sentence, searchWord) {
+  const items = sentence.split(' ')
+
+  for (const n in items) {
+    if (items[n].indexOf(searchWord) === 0) return Number(n) + 1
+  }
+
+  return -1
+}

src/string/defanging-an-ip-address.js

@@ -0,0 +1,7 @@
+/**
+ * @param {string} address
+ * @return {string}
+ */
+export const defangIPaddr = function (address) {
+  return address.replace(/\./g, '[.]')
+}

src/string/find-the-longest-substring-containing-vowels-in-even-counts.js

@@ -0,0 +1,30 @@
+/**
+ * @param {string} s
+ * @return {number}
+ */
+export const findTheLongestSubstring = function (s) {
+  const n = s.length
+  const pos = new Array(1 << 5).fill(-1)
+  let ans = 0; let status = 0
+  pos[0] = 0
+  for (let i = 0; i < n; ++i) {
+    const ch = s.charAt(i)
+    if (ch === 'a') {
+      status ^= 1 << 0
+    } else if (ch === 'e') {
+      status ^= 1 << 1
+    } else if (ch === 'i') {
+      status ^= 1 << 2
+    } else if (ch === 'o') {
+      status ^= 1 << 3
+    } else if (ch === 'u') {
+      status ^= 1 << 4
+    }
+    if (~pos[status]) {
+      ans = Math.max(ans, i + 1 - pos[status])
+    } else {
+      pos[status] = i + 1
+    }
+  }
+  return ans
+}

src/string/longest-common-prefix.js

@@ -0,0 +1,27 @@
+const comp = (left, right, res) => {
+  if (left.length > right.length) {
+    [left, right] = [right, left]
+  }
+
+  res = res || left
+  while (right.indexOf(res) !== 0 && res.length > 0) {
+    res = res.slice(0, res.length - 1)
+  }
+
+  return res
+}
+
+/**
+ * @param {string[]} strs
+ * @return {string}
+ */
+export const longestCommonPrefix = function (strs) {
+  if (strs.length < 2) return strs[0] || ''
+
+  let res = strs[0]
+  for (let i = 1; i < strs.length; i++) {
+    res = comp(res, strs[i])
+  }
+
+  return res
+}

src/string/longest-palindromic-substring.js

@@ -0,0 +1,36 @@
+export const longestPalindrome = function (s) {
+  if (!s || !s.trim()) return ''
+  if (s.length === 1) return s
+  if (s.length === 2) return s[0] === s[1] ? s[0] + s[1] : s[1]
+  let result = ''
+  /**
+  *扩散坐标
+  */
+  const calPalindromeIndex = function (left, right, s) {
+    const len = s.length
+    while (left >= 0 && right < len && s[left] === s[right]) {
+      left--
+      right++
+    }
+    return { left: left + 1, right: right }
+  }
+  for (let i = 0, len = s.length; i < len; i++) {
+    let even = ''
+    let odd = ''
+    if (s[i] === s[i + 1]) {
+      // 经过当前位与下一位判断已构成回文，扩散位直接从下一位开始，可以提速
+      const evenIndex = calPalindromeIndex(i - 1, i + 2, s)
+      even = s.slice(evenIndex.left, evenIndex.right)
+    }
+    const oddIndex = calPalindromeIndex(i - 1, i + 1, s)
+    odd = s.slice(oddIndex.left, oddIndex.right)
+    const re = odd.length > even.length ? odd : even
+    result = result.length > re.length ? result : re
+  }
+  return result
+}
+
+// 作者：liu-zi-qian-2
+// 链接：https://leetcode-cn.com/problems/longest-palindromic-substring/solution/5-zui-chang-hui-wen-zi-chuan-by-liu-zi-qian-2/
+// 来源：力扣（LeetCode）
+// 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

src/string/maximum-number-of-vowels-in-a-substring-of-given-length.js

@@ -0,0 +1,45 @@
+const v = ['a', 'e', 'i', 'o', 'u']
+const check = (str) => {
+  const map = new Map()
+
+  for (let i = 0; i < str.length; i++) {
+    if (v.includes(str[i])) {
+      map.set(str[i], (map.get(str[i]) || 0) + 1)
+    }
+  }
+
+  let res = 0
+  map.forEach(n => {
+    res += n
+  })
+  return res
+}
+
+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {number}
+ */
+export const maxVowels = function (s, k) {
+  let i = 0
+  let res = 0
+
+  const temp = s.slice(i, i + k)
+  let tmp = check(temp)
+  res = Math.max(res, tmp)
+
+  while (i + k <= s.length) {
+    i++
+    if (v.includes(s[i + k - 1])) {
+      tmp++
+    }
+
+    if (v.includes(s[i - 1])) {
+      tmp--
+    }
+
+    res = Math.max(res, tmp)
+  }
+
+  return res
+}

src/string/minimum-window-substring.js

@@ -0,0 +1,40 @@
+/**
+ * @param {string} s
+ * @param {string} t
+ * @return {string}
+ */
+export const minWindow = function (s, t) {
+  const sLen = s.length; const tLen = t.length
+  const tMap = new Map(); const map = new Map()
+  let l = 0; let r = 0; let res = ''
+
+  for (let i = 0; i < tLen; i++) {
+    tMap.set(t[i], (tMap.get(t[i]) || 0) + 1)
+  }
+
+  const check = (cMap) => {
+    const check = new Map(
+      [...cMap].filter(([k, v]) => tMap.has(k) && v >= tMap.get(k))
+    )
+
+    return check.size === tMap.size
+  }
+
+  while (r <= sLen) {
+    if (t.includes(s[r])) {
+      map.set(s[r], (map.get(s[r]) || 0) + 1)
+    }
+
+    while (check(map) && l <= r) {
+      const tmp = s.slice(l, r + 1)
+      if (res.length === 0 || tmp.length <= res.length) res = tmp
+      if (t.includes(s[l]) && map.has(s[l])) {
+        map.set(s[l], map.get(s[l]) - 1)
+      }
+      l++
+    }
+    r++
+  }
+
+  return res
+}

src/string/valid-palindrome-ii.js

@@ -0,0 +1,37 @@
+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+export const validPalindrome = function (s, flag = true) {
+  let l = 0; let r = s.length - 1
+  while (l < r && s[l] === s[r]) {
+    l++; r--
+  }
+  if (l >= r) return true // 说明是回文
+  if (flag) return validPalindrome(s.slice(l, r), false) || validPalindrome(s.slice(l + 1, r + 1), false) // 如果不是回文，切左边或者右边，再判断一次
+  return false
+}
+
+// 暴力解法，会超时
+// /**
+//  * @param {string} s
+//  * @return {boolean}
+//  */
+// export const validPalindrome = function (s) {
+//   const len = s.length
+//   const tmpLen = len - 1
+//   let mid = len >> 1
+//   if (s.substring(0, mid) === s.substring(len % 2 ? mid + 1 : mid, len).split('').reverse().join('')) {
+//     return true
+//   }
+
+//   mid = tmpLen >> 1
+//   for (let n = 0; n < len; n++) {
+//     const tmp = s.substring(0, n) + s.substring(n + 1, len)
+//     if (tmp.substring(0, mid) === tmp.substring(tmpLen % 2 ? mid + 1 : mid, tmpLen).split('').reverse().join('')) {
+//       return true
+//     }
+//   }
+
+//   return false
+// }

src/tree/binary-tree-level-order-traversal-ii.js

@@ -0,0 +1,28 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number[][]}
+ */
+export const levelOrder = function (root) {
+  const res = []
+  const queue = [root]
+  while (queue.length) { // BFS
+    const tmp = []
+    const leave = queue.length // 记录这一层有几个
+    for (let i = 0; i < leave; i++) { // 一次性把固定个数的队列执行完
+      const node = queue.shift()
+      if (node && node.left) queue.push(node.left)
+      if (node && node.right) queue.push(node.right)
+      if (node) tmp.push(node.val)
+    }
+    if (tmp.length) res.unshift(tmp)
+  }
+
+  return res
+}

src/tree/binary-tree-level-order-traversal.js

@@ -0,0 +1,28 @@
+/**
+ * Definition for a binary tree node.
+ * function TreeNode(val) {
+ *     this.val = val;
+ *     this.left = this.right = null;
+ * }
+ */
+/**
+ * @param {TreeNode} root
+ * @return {number[][]}
+ */
+export const levelOrder = function (root) {
+  const res = []
+  const queue = [root]
+  while (queue.length) { // BFS
+    const tmp = []
+    const leave = queue.length // 记录这一层有几个
+    for (let i = 0; i < leave; i++) { // 一次性把固定个数的队列执行完
+      const node = queue.shift()
+      if (node && node.left) queue.push(node.left)
+      if (node && node.right) queue.push(node.right)
+      if (node) tmp.push(node.val)
+    }
+    if (tmp.length) res.push(tmp)
+  }
+
+  return res
+}

src/tree/construct-binary-tree-from-preorder-and-inorder-traversal.js

@@ -0,0 +1,26 @@
+/**
+ * Definition for a binary tree node.
+ */
+function TreeNode (val) {
+  this.val = val
+  this.left = this.right = null
+}
+
+// 前序遍历：根左右
+// 中序遍历：左根右
+
+// 前序遍历知道根、中序遍历知道左边长度。
+
+/**
+ * @param {number[]} preorder
+ * @param {number[]} inorder
+ * @return {TreeNode}
+ */
+export const buildTree = function (preorder, inorder) {
+  if (inorder.length === 0) return null
+  const root = new TreeNode(preorder[0])
+  const index = inorder.indexOf(preorder[0])
+  root.left = buildTree(preorder.slice(1, index + 1), inorder.slice(0, index))
+  root.right = buildTree(preorder.slice(index + 1), inorder.slice(index + 1))
+  return root
+}

src/tree/convert-sorted-array-to-binary-search-tree.js

@@ -0,0 +1,20 @@
+import TreeNode from '../../test/tree/TreeNode'
+
+/**
+ * @param {number[]} nums
+ * @return {TreeNode}
+ */
+export const sortedArrayToBST = function (nums) {
+  if (!nums.length) return null
+
+  const creatTree = (left, right) => {
+    if (left > right) return null
+    const mid = Math.floor((left + right) / 2)
+    const root = new TreeNode(nums[mid])
+    root.left = creatTree(left, mid - 1)
+    root.right = creatTree(mid + 1, right)
+    return root
+  }
+
+  return creatTree(0, nums.length - 1)
+}

src/tree/unique-binary-search-trees.js

@@ -0,0 +1,12 @@
+// 卡塔兰数
+/**
+ * @param {number} n
+ * @return {number}
+ */
+export const numTrees = function (n) {
+  let C = 1
+  for (let i = 0; i < n; ++i) {
+    C = C * 2 * (2 * i + 1) / (i + 2)
+  }
+  return C
+}

test/array/daily-temperatures.test.js

@@ -0,0 +1,5 @@
+import { dailyTemperatures } from '../../src/array/daily-temperatures.js'
+
+test('每日温度', () => {
+  expect(dailyTemperatures([73, 74, 75, 71, 69, 72, 76, 73])).toEqual([1, 1, 4, 2, 1, 1, 0, 0])
+})

test/array/find-the-duplicate-number.test.js

@@ -0,0 +1,6 @@
+import { findDuplicate } from '../../src/array/find-the-duplicate-number'
+
+test('寻找重复数', () => {
+  expect(findDuplicate([1, 3, 4, 2, 2])).toBe(2)
+  expect(findDuplicate([3, 1, 3, 4, 2])).toBe(3)
+})

test/array/first-unique-character-in-a-string.test.js

@@ -0,0 +1,5 @@
+import { firstUniqChar } from '../../src/array/first-unique-character-in-a-string'
+
+test('第一个只出现一次的字符', () => {
+  expect(firstUniqChar('leetcode')).toBe('l')
+})

test/array/first-unique-number-in-data-stream.test.js

@@ -0,0 +1,7 @@
+import { firstUniqueNumber } from '../../src/array/first-unique-number-in-data-stream'
+
+test('数据流中第一个唯一的数字', () => {
+  expect(firstUniqueNumber([1, 2, 2, 1, 3, 4, 4, 5, 6], 5)).toBe(3)
+  expect(firstUniqueNumber([1, 2, 2, 1, 3, 4, 4, 5, 6], 7)).toBe(-1)
+  expect(firstUniqueNumber([1, 2, 2, 1, 3, 4], 3)).toBe(3)
+})

test/array/highest-frequency-ip.test.js

@@ -0,0 +1,6 @@
+import { highestFrequency } from '../../src/array/highest-frequency-ip'
+
+test('最高频率的IP', () => {
+  expect(highestFrequency(['192.168.1.1', '192.118.2.1', '192.168.1.1'])).toEqual('192.168.1.1')
+  expect(highestFrequency(['192.168.1.1', '192.118.2.1', '192.168.1.1', '192.118.2.1', '192.118.2.1'])).toEqual('192.118.2.1')
+})

test/array/house-robber.test.js

@@ -0,0 +1,7 @@
+import { rob } from '../../src/array/house-robber'
+
+test('打家劫舍', () => {
+  expect(rob([1, 2, 3, 1])).toBe(4)
+  expect(rob([2, 7, 9, 3, 1])).toBe(12)
+  expect(rob([828, 125, 740, 724, 983, 321, 773, 678, 841, 842, 875, 377, 674, 144, 340, 467, 625, 916, 463, 922, 255, 662, 692, 123, 778, 766, 254, 559, 480, 483, 904, 60, 305, 966, 872, 935, 626, 691, 832, 998, 508, 657, 215, 162, 858, 179, 869, 674, 452, 158, 520, 138, 847, 452, 764, 995, 600, 568, 92, 496, 533, 404, 186, 345, 304, 420, 181, 73, 547, 281, 374, 376, 454, 438, 553, 929, 140, 298, 451, 674, 91, 531, 685, 862, 446, 262, 477, 573, 627, 624, 814, 103, 294, 388])).toBe(29123)
+})

test/array/kids-with-the-greatest-number-of-candies.test.js

@@ -0,0 +1,7 @@
+import { kidsWithCandies } from '../../src/array/kids-with-the-greatest-number-of-candies'
+
+test('拥有最多糖果的孩子', () => {
+  expect(kidsWithCandies([2, 3, 5, 1, 3], 3)).toEqual([true, true, true, false, true])
+  expect(kidsWithCandies([4, 2, 1, 1, 2], 1)).toEqual([true, false, false, false, false])
+  expect(kidsWithCandies([12, 1, 12], 10)).toEqual([true, false, true])
+})

test/array/kth-smallest-element-in-a-sorted-matrix.test.js

@@ -0,0 +1,20 @@
+import { kthSmallest } from '../../src/array/kth-smallest-element-in-a-sorted-matrix.js'
+
+test('有序矩阵中第K小的元素', () => {
+  expect(kthSmallest([
+    [1, 5, 9],
+    [10, 11, 13],
+    [12, 13, 15]
+  ], 8)).toBe(13)
+
+  expect(kthSmallest([
+    [1, 5, 7],
+    [3, 7, 8],
+    [4, 8, 9]
+  ], 4)).toBe(5)
+
+  expect(kthSmallest([
+    [1, 2],
+    [3, 4]
+  ], 3)).toBe(3)
+})

test/array/lemonade-change.test.js

@@ -0,0 +1,9 @@
+import { lemonadeChange } from '../../src/array/lemonade-change'
+
+test('柠檬水找零', () => {
+  expect(lemonadeChange([5, 5, 5, 5, 10, 5, 10, 10, 10, 20])).toBe(true)
+  expect(lemonadeChange([5, 5, 5, 10, 20])).toBe(true)
+  expect(lemonadeChange([5, 5, 10])).toBe(true)
+  expect(lemonadeChange([10, 10])).toBe(false)
+  expect(lemonadeChange([5, 5, 10, 10, 20])).toBe(false)
+})

test/array/lru-cache.test.js

@@ -0,0 +1,15 @@
+import LRUCache from '../../src/array/lru-cache'
+
+test('LRU缓存机制', () => {
+  const cache = new LRUCache(2 /* 缓存容量 */)
+
+  cache.put(1, 1)
+  cache.put(2, 2)
+  expect(cache.get(1)).toBe(1) // 返回  1
+  cache.put(3, 3) // 该操作会使得密钥 2 作废
+  expect(cache.get(2)).toBe(-1) // 返回 -1 (未找到)
+  cache.put(4, 4) // 该操作会使得密钥 1 作废
+  expect(cache.get(1)).toBe(-1) // 返回 -1 (未找到)
+  expect(cache.get(3)).toBe(3) // 返回  3
+  expect(cache.get(4)).toBe(4) // 返回  4
+})

test/array/maximum-length-of-repeated-subarray.test.js

@@ -0,0 +1,5 @@
+import { findLength } from '../../src/array/maximum-length-of-repeated-subarray.js'
+
+test('最长重复子数组', () => {
+  expect(findLength([1, 2, 3, 2, 1], [3, 2, 1, 4, 7])).toBe(3)
+})

test/array/maximum-product-subarray.test.js

@@ -0,0 +1,6 @@
+import { maxProduct } from '../../src/array/maximum-product-subarray'
+
+test('', () => {
+  expect(maxProduct([2, 3, -2, 4])).toBe(6)
+  expect(maxProduct([-2, 0, -1])).toBe(0)
+})

test/array/product-of-array-except-self.test.js

@@ -0,0 +1,5 @@
+import { productExceptSelf } from '../../src/array/product-of-array-except-self'
+
+test('除自身以外数组的乘积', () => {
+  expect(productExceptSelf([1, 2, 3, 4])).toEqual([24, 12, 8, 6])
+})

test/array/subarray-sums-divisible-by-k.test.js

@@ -0,0 +1,5 @@
+import { subarraysDivByK } from '../../src/array/subarray-sums-divisible-by-k'
+
+test('和可被K整除的子数组', () => {
+  expect(subarraysDivByK([4, 5, 0, -2, -3, 1], 5)).toBe(7)
+})

test/graphs/course-schedule-ii.test.js

@@ -0,0 +1,6 @@
+import { findOrder } from '../../src/graphs/course-schedule-ii'
+
+test('课程表 II', () => {
+  expect(findOrder(2, [[1, 0]])).toEqual([0, 1])
+  expect(findOrder(4, [[1, 0], [2, 0], [3, 1], [3, 2]])).toEqual([0, 1, 2, 3])
+})

test/list/ListNode.js

@@ -15,3 +15,13 @@ export const arrToList = (arr) => {
 
   return head
 }
+
+export const listToArr = (list) => {
+  const arr = []
+  while (list) {
+    arr.push(list.val)
+    list = list.next
+  }
+
+  return arr
+}

test/list/reverse-nodes-in-k-group.test.js

@@ -0,0 +1,11 @@
+import { reverseKGroup } from '../../src/list/reverse-nodes-in-k-group'
+import { arrToList, listToArr } from './ListNode'
+
+test('K个一组翻转链表', () => {
+  const res = reverseKGroup(arrToList([1, 2, 3, 4, 5]), 3)
+  expect(listToArr(res)).toEqual([3, 2, 1, 4, 5])
+  expect(res).toEqual(arrToList([3, 2, 1, 4, 5]))
+  expect(reverseKGroup(arrToList([1, 2, 3, 4, 5]), 2)).toEqual(arrToList([2, 1, 4, 3, 5]))
+  expect(reverseKGroup(arrToList([1, 2, 3, 4, 5, 6]), 2)).toEqual(arrToList([2, 1, 4, 3, 6, 5]))
+  expect(reverseKGroup(arrToList([1, 2, 3, 4, 5, 6, 7, 8]), 2)).toEqual(arrToList([2, 1, 4, 3, 6, 5, 8, 7]))
+})

test/math/3sum.test.js

@@ -0,0 +1,8 @@
+import { threeSum } from '../../src/math/3sum.js'
+
+test('三数之和', () => {
+  expect(threeSum([-1, 0, 1, 2, -1, -4])).toEqual([
+    [-1, -1, 2],
+    [-1, 0, 1]
+  ])
+})

test/math/add-binary.test.js

@@ -0,0 +1,6 @@
+import { addBinary } from '../../src/math/add-binary'
+
+test('二进制求和', () => {
+  expect(addBinary('11', '1')).toEqual('100')
+  expect(addBinary('1010', '1011')).toEqual('10101')
+})

test/math/median-of-two-sorted-arrays.test.js

@@ -0,0 +1,7 @@
+import { findMedianSortedArrays } from '../../src/math/median-of-two-sorted-arrays'
+
+test('寻找两个正序数组的中位数', () => {
+  expect(findMedianSortedArrays([], [1])).toBe(1.0)
+  expect(findMedianSortedArrays([1, 3], [2])).toBe(2.0)
+  expect(findMedianSortedArrays([1, 2], [3, 4])).toBe(2.5)
+})

test/math/perfect-number.test.js

@@ -0,0 +1,6 @@
+import { checkPerfectNumber } from '../../src/math/perfect-number'
+
+test('完美数', () => {
+  expect(checkPerfectNumber(28)).toBe(true)
+  expect(checkPerfectNumber(2)).toBe(false)
+})

test/math/subarray-sum-equals-k.test.js

@@ -0,0 +1,5 @@
+import { subarraySum } from '../../src/math/subarray-sum-equals-k'
+
+test('和为K的子数组', () => {
+  expect(subarraySum([1, 1, 1], 2)).toBe(2)
+})

test/stack/decode-string.test.js

@@ -0,0 +1,7 @@
+import { decodeString } from '../../src/stack/decode-string'
+
+test('字符串解码', () => {
+  expect(decodeString('3[a]2[bc]')).toBe('aaabcbc')
+  expect(decodeString('3[a2[c]]')).toBe('accaccacc')
+  expect(decodeString('2[abc]3[cd]ef')).toBe('abcabccdcdcdef')
+})

test/stack/yong-liang-ge-zhan-shi-xian-dui-lie-lcof.test.js

@@ -0,0 +1,67 @@
+import { CQueue } from '../../src/stack/yong-liang-ge-zhan-shi-xian-dui-lie-lcof.js'
+
+test('用两个栈实现队列 - 1', () => {
+  let obj = null
+  const action = ['CQueue', 'appendTail', 'deleteHead', 'deleteHead']
+  const queue = [[], [3], [], []]
+  const res = []
+  while (queue.length) {
+    const tmp = action.shift()
+    const val = queue.shift()
+    if (tmp === 'CQueue') {
+      obj = new CQueue()
+      res.push(null)
+    } else if (tmp === 'appendTail') {
+      obj.appendTail(val[0])
+      res.push(null)
+    } else if (tmp === 'deleteHead') {
+      res.push(obj.deleteHead())
+    }
+  }
+
+  expect(res).toEqual([null, null, 3, -1])
+})
+
+test('用两个栈实现队列 - 2', () => {
+  let obj = null
+  const action = ['CQueue', 'deleteHead', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead']
+  const queue = [[], [], [5], [2], [], []]
+  const res = []
+  while (queue.length) {
+    const tmp = action.shift()
+    const val = queue.shift()
+    if (tmp === 'CQueue') {
+      obj = new CQueue()
+      res.push(null)
+    } else if (tmp === 'appendTail') {
+      obj.appendTail(val[0])
+      res.push(null)
+    } else if (tmp === 'deleteHead') {
+      res.push(obj.deleteHead())
+    }
+  }
+
+  expect(res).toEqual([null, -1, null, null, 5, 2])
+})
+
+test('用两个栈实现队列 - 3', () => {
+  let obj = null
+  const action = ['CQueue', 'deleteHead', 'appendTail', 'deleteHead', 'deleteHead', 'deleteHead', 'deleteHead', 'appendTail', 'deleteHead', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'deleteHead', 'appendTail', 'deleteHead', 'deleteHead', 'deleteHead', 'deleteHead', 'appendTail', 'appendTail', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'deleteHead', 'appendTail', 'appendTail', 'deleteHead', 'appendTail', 'deleteHead', 'appendTail', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'deleteHead', 'deleteHead', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'appendTail', 'deleteHead', 'appendTail', 'appendTail', 'appendTail', 'appendTail', 'deleteHead', 'appendTail', 'deleteHead', 'deleteHead', 'appendTail', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'appendTail', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'deleteHead', 'deleteHead', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'appendTail', 'deleteHead', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'appendTail', 'deleteHead', 'appendTail', 'appendTail', 'appendTail', 'deleteHead', 'appendTail', 'appendTail', 'appendTail', 'appendTail', 'deleteHead', 'deleteHead', 'deleteHead', 'appendTail', 'deleteHead', 'appendTail', 'deleteHead', 'appendTail', 'appendTail']
+  const queue = [[], [], [97], [], [], [], [], [15], [], [1], [43], [], [], [], [18], [], [], [], [], [36], [69], [21], [91], [], [], [22], [40], [], [], [], [81], [65], [], [77], [], [63], [96], [5], [], [], [35], [90], [], [], [], [], [77], [83], [], [], [52], [], [2], [66], [87], [90], [], [2], [], [], [33], [16], [72], [], [], [14], [78], [8], [], [], [], [], [3], [83], [], [], [13], [], [79], [44], [], [], [33], [], [55], [76], [12], [], [91], [24], [49], [47], [], [], [], [85], [], [69], [], [94], [52]]
+  const res = []
+  while (queue.length) {
+    const tmp = action.shift()
+    const val = queue.shift()
+    if (tmp === 'CQueue') {
+      obj = new CQueue()
+      res.push(null)
+    } else if (tmp === 'appendTail') {
+      obj.appendTail(val[0])
+      res.push(null)
+    } else if (tmp === 'deleteHead') {
+      res.push(obj.deleteHead())
+    }
+  }
+
+  expect(res).toEqual([null, -1, null, 97, -1, -1, -1, null, 15, null, null, 1, 43, -1, null, 18, -1, -1, -1, null, null, null, null, 36, 69, null, null, 21, 91, 22, null, null, 40, null, 81, null, null, null, 65, 77, null, null, 63, 96, 5, 35, null, null, 90, 77, null, 83, null, null, null, null, 52, null, 2, 66, null, null, null, 87, 90, null, null, null, 2, 33, 16, 72, null, null, 14, 78, null, 8, null, null, 3, 83, null, 13, null, null, null, 79, null, null, null, null, 44, 33, 55, null, 76, null, 12, null, null])
+})

test/string/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence.test.js

@@ -0,0 +1,5 @@
+import { isPrefixOfWord } from '../../src/string/check-if-a-word-occurs-as-a-prefix-of-any-word-in-a-sentence'
+
+test('检查单词是否为句中其他单词的前缀', () => {
+  expect(isPrefixOfWord('i love eating burger', 'burg')).toBe(4)
+})

test/string/defanging-an-ip-address.test.js

@@ -0,0 +1,6 @@
+import { defangIPaddr } from '../../src/string/defanging-an-ip-address'
+
+test('IP地址无效化', () => {
+  expect(defangIPaddr('1.1.1.1')).toBe('1[.]1[.]1[.]1')
+  expect(defangIPaddr('255.100.50.0')).toBe('255[.]100[.]50[.]0')
+})

test/string/find-the-longest-substring-containing-vowels-in-even-counts.test.js

@@ -0,0 +1,7 @@
+import { findTheLongestSubstring } from '../../src/string/find-the-longest-substring-containing-vowels-in-even-counts'
+
+test('每个元音包含偶数次的最长子字符串', () => {
+  expect(findTheLongestSubstring('eleetminicoworoep')).toBe(13)
+  expect(findTheLongestSubstring('leetcodeisgreat')).toBe(5)
+  expect(findTheLongestSubstring('bcbcbc')).toBe(6)
+})

test/string/longest-common-prefix.test.js

@@ -0,0 +1,8 @@
+import { longestCommonPrefix } from '../../src/string/longest-common-prefix'
+
+test('最长公共前缀', () => {
+  expect(longestCommonPrefix(['flower', 'flow', 'flight'])).toBe('fl')
+  expect(longestCommonPrefix(['dog', 'racecar', 'car'])).toBe('')
+  expect(longestCommonPrefix(['a'])).toBe('a')
+  expect(longestCommonPrefix(['caa', '', 'a', 'acb'])).toBe('')
+})

test/string/longest-palindromic-substring.test.js

@@ -0,0 +1,6 @@
+import { longestPalindrome } from '../../src/string/longest-palindromic-substring'
+
+test('最长回文字符串', () => {
+  expect(longestPalindrome('babad')).toBe('aba') // bab也是有效答案
+  expect(longestPalindrome('cbbd')).toBe('bb')
+})

test/string/maximum-number-of-vowels-in-a-substring-of-given-length.test.js

@@ -0,0 +1,6 @@
+import { maxVowels } from '../../src/string/maximum-number-of-vowels-in-a-substring-of-given-length'
+
+test('定长子串中元音的最大数目', () => {
+  expect(maxVowels('abciiidef', 3)).toBe(3)
+  expect(maxVowels('rhythms', 4)).toBe(0)
+})

test/string/minimum-window-substring.test.js

@@ -0,0 +1,9 @@
+import { minWindow } from '../../src/string/minimum-window-substring'
+
+test('最小覆盖子串', () => {
+  expect(minWindow('ab', 'a')).toBe('a')
+  expect(minWindow('ADOBECODEBANC', 'ABC')).toBe('BANC')
+  expect(minWindow('a', 'a')).toBe('a')
+  expect(minWindow('a', 'aa')).toBe('')
+  expect(minWindow('abcdc', 'da')).toBe('abcd')
+})

test/string/valid-palindrome-ii.test.js

@@ -0,0 +1,8 @@
+import { validPalindrome } from '../../src/string/valid-palindrome-ii'
+
+test('验证回文字符串 Ⅱ', () => {
+  expect(validPalindrome('aba')).toBe(true)
+  expect(validPalindrome('abba')).toBe(true)
+  expect(validPalindrome('abca')).toBe(true)
+  expect(validPalindrome('abcda')).toBe(false)
+})

test/tree/binary-tree-level-order-traversal-ii.test.js

@@ -0,0 +1,14 @@
+import { levelOrder } from '../../src/tree/binary-tree-level-order-traversal-ii'
+import Tree from './Tree.js'
+
+test('二叉树的层序遍历 II', () => {
+  const source = [3, 9, 20, null, null, 15, 7]
+  expect(levelOrder(Tree.arrToTree(source))).toEqual([
+    [15, 7],
+    [9, 20],
+    [3]
+  ])
+
+  const source1 = []
+  expect(levelOrder(Tree.arrToTree(source1))).toEqual([])
+})

test/tree/binary-tree-level-order-traversal.test.js

@@ -0,0 +1,14 @@
+import { levelOrder } from '../../src/tree/binary-tree-level-order-traversal'
+import Tree from './Tree.js'
+
+test('二叉树的层序遍历', () => {
+  const source = [3, 9, 20, null, null, 15, 7]
+  expect(levelOrder(Tree.arrToTree(source))).toEqual([
+    [3],
+    [9, 20],
+    [15, 7]
+  ])
+
+  const source1 = []
+  expect(levelOrder(Tree.arrToTree(source1))).toEqual([])
+})

test/tree/construct-binary-tree-from-preorder-and-inorder-traversal.test.js

@@ -0,0 +1,5 @@
+import { buildTree } from '../../src/tree/construct-binary-tree-from-preorder-and-inorder-traversal'
+
+test('从前序与中序遍历序列构造二叉树', () => {
+  expect(buildTree([3, 9, 20, 15, 7], [9, 3, 15, 20, 7])).toEqual({ left: { left: null, right: null, val: 9 }, right: { left: { left: null, right: null, val: 15 }, right: { left: null, right: null, val: 7 }, val: 20 }, val: 3 })
+})

test/tree/convert-sorted-array-to-binary-search-tree.test.js

@@ -0,0 +1,5 @@
+import { sortedArrayToBST } from '../../src/tree/convert-sorted-array-to-binary-search-tree.js'
+
+test('将有序数组转换为二叉搜索树', () => {
+  expect(sortedArrayToBST([-10, -3, 0, 5, 9])).toEqual({ left: { left: null, right: { left: null, right: null, val: -3 }, val: -10 }, right: { left: null, right: { left: null, right: null, val: 9 }, val: 5 }, val: 0 })
+})

test/tree/unique-binary-search-trees.test.js

@@ -0,0 +1,5 @@
+import { numTrees } from '../../src/tree/unique-binary-search-trees'
+
+test('不同的二叉搜索树', () => {
+  expect(numTrees(3)).toBe(5)
+})