add: 只出现一次的数字等

This commit is contained in:
yi-ge 2019-03-10 16:43:04 +08:00 committed by yi-ge
parent 930ff1be11
commit 8239e7afd5
106 changed files with 7978 additions and 0 deletions

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.babelrc Normal file
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{
"presets": ["@babel/preset-env"]
}

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module.exports = {
root: true,
parserOptions: {
parser: 'babel-eslint',
ecmaVersion: 2017,
sourceType: 'module'
},
env: {
browser: true,
node: true
},
extends: ['standard', 'plugin:jest/recommended'],
rules: {
// allow async-await
'generator-star-spacing': 'off',
// allow debugger during development
'no-debugger': process.env.NODE_ENV === 'production' ? 'error' : 'off'
}
}

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# This workflow will do a clean install of node dependencies, build the source code and run tests across different versions of node
# For more information see: https://help.github.com/actions/language-and-framework-guides/using-nodejs-with-github-actions
name: Node.js CI
on:
push:
branches: [ master ]
pull_request:
branches: [ master ]
jobs:
build:
runs-on: ubuntu-latest
strategy:
matrix:
node-version: [10.x, 12.x]
steps:
- uses: actions/checkout@v2
- name: Use Node.js ${{ matrix.node-version }}
uses: actions/setup-node@v1
with:
node-version: ${{ matrix.node-version }}
- run: npm install -g yarn
- run: yarn install
- run: yarn test
env:
CI: true

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tasks:
- init: yarn install

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language: node_js
node_js:
- 13
script:
- yarn test
- yarn coverage:ci

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{
"cSpell.words": [
"lcci",
"lcof",
"nums",
"zhong"
]
}

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# JavaScript 算法练习
[![license](https://img.shields.io/github/license/yi-ge/js-practice.svg?style=flat-square)](https://github.com/yi-ge/js-practice/blob/master/LICENSE)
![Travis (.org)](https://img.shields.io/travis/yi-ge/js-practice?style=flat-square)
![Coveralls github](https://img.shields.io/coveralls/github/yi-ge/js-practice?style=flat-square)
[![GitHub last commit](https://img.shields.io/github/last-commit/yi-ge/js-practice.svg?style=flat-square)](https://github.com/yi-ge/js-practice)
[![Gitpod Ready-to-Code](https://img.shields.io/badge/Gitpod-Ready--to--Code-blue?logo=gitpod&style=flat-square)](https://gitpod.io/#https://github.com/yi-ge/js-practice)
![Node.js CI](https://github.com/yi-ge/js-practice/workflows/Node.js%20CI/badge.svg)
[![JavaScript Style Guide](https://cdn.rawgit.com/standard/standard/master/badge.svg)](https://github.com/standard/standard)
## 字符串
- [反转字符串中的单词](src/string/reverse-words-in-a-string.js)
- LeetCode 557. 反转字符串中的单词 III https://leetcode-cn.com/problems/reverse-words-in-a-string-iii/
- LintCode 1173. 反转字符串 III https://www.lintcode.com/problem/reverse-words-in-a-string-iii/description
- [计数二进制子串](src/string/count-binary-substrings.js)
- LeetCode 696. 计数二进制子串 https://leetcode-cn.com/problems/count-binary-substrings/
- LintCode 1079. 连续子串计数 https://www.lintcode.com/problem/count-binary-substrings/description
- [重复的子串](src/string/repeated-substring-pattern.js)
- LeetCode 459. 重复的子字符串 https://leetcode-cn.com/problems/repeated-substring-pattern/
- LintCode 1227. 重复的子串模式 https://www.lintcode.com/problem/repeated-substring-pattern/description
- [正则表达式匹配](src/string/regular-expression-matching.js)
- LeetCode 10. 正则表达式匹配 https://leetcode-cn.com/problems/regular-expression-matching/
- LintCode 154. 正则表达式匹配 https://www.lintcode.com/problem/regular-expression-matching/description
- [恢复 IP 地址](src/string/restore-ip-addresses.js)
- LeetCode 93. 复原 IP 地址 https://leetcode-cn.com/problems/restore-ip-addresses/
- LintCode 426. 恢复 IP 地址 https://www.lintcode.com/problem/restore-ip-addresses/description
- [存在重复元素](src/string/contains-duplicate.js)
- LeetCode 217. 存在重复元素 https://leetcode-cn.com/problems/contains-duplicate/
- [判断字符串是否没有重复字符](src/string/unique-characters.js)
- LintCode 157. 判断字符串是否没有重复字符 https://www.lintcode.com/problem/unique-characters/description
- [比较字符串](src/string/compare-strings.js)
- LintCode 55. 比较字符串 https://www.lintcode.com/problem/compare-strings/description
## 数组
- [电话号码的字母组合](src/array/letter-combinations-of-a-phone-number.js)
- LeetCode 17. 电话号码的字母组合 https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/
- LintCode 425. 电话号码的字母组合 https://www.lintcode.com/problem/letter-combinations-of-a-phone-number/description
- [卡牌分组](src/array/x-of-a-kind-in-a-deck-of-cards.js)
- LeetCode 914. 卡牌分组 https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards/
- [删除排序数组中的重复项](src/array/remove-duplicates-from-sorted-array.js)
- LeetCode 26. 删除排序数组中的重复项 https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/
- LintCode 100. 删除排序数组中的重复数字 https://www.lintcode.com/problem/remove-duplicates-from-sorted-array/description
- [能否种花](src/array/can-place-flowers.js)
- LeetCode 605. 种花问题 https://leetcode-cn.com/problems/can-place-flowers/
- LintCode 1138. 能否放置花 https://www.lintcode.com/problem/can-place-flowers/description
- [循环升序数组](src/array/loop-asc-array.js)
- 假设有一个升序数组,经过不确定长度的偏移,得到一个新的数组,我们称为循环升序数组。
(例如:[0, 3, 4, 6, 7] 可能变成 [6, 7, 0, 3, 4]
给定一个数字和一个循环升序数组,判断这个数字是否在这个数组内,在的话返回 true否则返回 false。要求时间复杂度为 O(logN)
示例 1
输入nums = [6, 7, 0, 3, 4] target = 0
输出true
示例 2
输入nums = [6, 7, 0, 3, 4] target = 5
输出false
- [循环数组中的环-暂无解](src/array/circular-array-loop.js)
- LeetCode 457. 环形数组循环 https://leetcode-cn.com/problems/circular-array-loop/
- LintCode 1229. 循环数组中的环 https://www.lintcode.com/problem/circular-array-loop/description
- [格雷编码](src/array/gray-code.js)
- LeetCode 89. 格雷编码 https://leetcode-cn.com/problems/gray-code/
- LintCode 411. 格雷编码 https://www.lintcode.com/problem/gray-code/description
- [数组划分](src/array/partition-array.js)
- LintCode 31. 数组划分 https://www.lintcode.com/problem/partition-array/description
- [冒泡排序](src/array/bubble-sort.js)
- [选择排序](src/array/select-sort.js)
- [按奇偶排序数组](src/array/sort-array-by-parity.js)
- LeetCode 922. 按奇偶排序数组 II https://leetcode-cn.com/problems/sort-array-by-parity-ii/
- [数组中的第 K 个最大元素](src/array/kth-largest-element-in-an-array.js)
- LeetCode 215. 数组中的第 K 个最大元素 https://leetcode-cn.com/problems/kth-largest-element-in-an-array/
- LeetCode 414. 第三大的数【相似,需去重】 https://leetcode-cn.com/problems/third-maximum-number/submissions/
- LintCode 5. 第 k 大元素 https://www.lintcode.com/problem/kth-largest-element/description
- LintCode 606. 第 K 大的元素 II https://www.lintcode.com/problem/kth-largest-element-ii/description
- LintCode 544. 前 K 大数【相似】 https://www.lintcode.com/problem/top-k-largest-numbers/description
- LintCode 479. 数组第二大数【相似】 https://www.lintcode.com/problem/second-max-of-array/description
- [最大间距](src/array/maximum-gap.js)
- LeetCode 164. 最大间距 https://leetcode-cn.com/problems/maximum-gap/
- LintCode 400. 最大间距 https://www.lintcode.com/problem/maximum-gap/
- [缺失的第一个正数](src/array/first-missing-positive.js)
- LeetCode 41. 缺失的第一个正数 https://leetcode-cn.com/problems/first-missing-positive/
- LintCode 189. 丢失的第一个正整数 https://www.lintcode.com/problem/first-missing-positive/description
- LintCode 681. 缺失的第一个素数【相似】 https://www.lintcode.com/problem/first-missing-prime-number/description
- [缺失的第一个素数](src/array/first-missing-prime-number.js)
- LintCode 681. 缺失的第一个素数 https://www.lintcode.com/problem/first-missing-prime-number/description
- [串联所有单词的子串](src/array/substring-with-concatenation-of-all-words.js)
- LeetCode 30. 串联所有单词的子串 https://leetcode-cn.com/problems/substring-with-concatenation-of-all-words/
- LintCode 1362. 包含所有单词连接的子串 https://www.lintcode.com/problem/substring-with-concatenation-of-all-words/description
- [买卖股票的最佳时机 II](src/array/best-time-to-buy-and-sell-stock-ii.js)
- LeetCode 122. 买卖股票的最佳时机 II https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-ii/
- LintCode 150. 买卖股票的最佳时机 II https://www.lintcode.com/problem/best-time-to-buy-and-sell-stock-ii/description
- [搜索插入位置](src/array/search-insert-position.js)
- LeetCode 35. 搜索插入位置 https://leetcode-cn.com/problems/search-insert-position/
- LintCode 20. 搜索插入位置 https://www.lintcode.com/problem/search-insert-position/description
- [二分查找](src/array/binary-search.js)
- LeetCode 35. 二分查找 https://leetcode-cn.com/problems/binary-search/
- LintCode 20. 二分查找 https://www.lintcode.com/problem/first-position-of-target/description
- [查找常用字符](src/array/find-common-characters.js)
- LeetCode 1002. 查找常用字符 https://leetcode-cn.com/problems/find-common-characters/
- [01 矩阵](src/array/01-matrix.js)
- LeetCode 542. 01 矩阵 https://leetcode-cn.com/problems/01-matrix/
- LintCode 974. 01 矩阵 https://www.lintcode.com/problem/01-matrix/description
- [合并区间](src/array/merge-intervals.js)
- LeetCode 56. 合并区间 https://leetcode-cn.com/problems/merge-intervals/
- LintCode 156. 合并区间 https://www.lintcode.com/problem/merge-intervals/description
- [跳跃游戏](src/array/jump-game.js)
- LeetCode 55. 跳跃游戏 https://leetcode-cn.com/problems/jump-game/
- LintCode 116. 跳跃游戏 https://www.lintcode.com/problem/jump-game/description
- [盛最多水的容器](src/array/container-with-most-water.js)
- LeetCode 11. 盛最多水的容器 https://leetcode-cn.com/problems/container-with-most-water/
- LintCode 383. 装最多水的容器 https://www.lintcode.com/problem/container-with-most-water/description
- [统计重复个数](src/array/count-the-repetitions.js)
- LeetCode 466. 统计重复个数 https://leetcode-cn.com/problems/count-the-repetitions/
- LintCode 1224. 统计重复个数 https://www.lintcode.com/problem/count-the-repetitions/description
- [岛屿的个数](src/array/number-of-islands.js)
- LeetCode 200. 岛屿数量 https://leetcode-cn.com/problems/number-of-islands/
- LintCode 433. 岛屿的个数 https://www.lintcode.com/problem/number-of-islands/description
- [统计「优美子数组」](src/array/count-number-of-nice-subarrays.js)
- LeetCode 1248. 统计「优美子数组」 https://leetcode-cn.com/problems/count-number-of-nice-subarrays/
- [主元素](src/array/majority-element.js)
- LeetCode 面试题 17.10. 主要元素 https://leetcode-cn.com/problems/find-majority-element-lcci/
- LintCode 46. 主元素 https://www.lintcode.com/problem/majority-element/description
- [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/
- 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)
- LeetCode 33. 搜索旋转排序数组 https://leetcode-cn.com/problems/search-in-rotated-sorted-array/
- LintCode 62. 搜索旋转排序数组 https://www.lintcode.com/problem/search-in-rotated-sorted-array/description
- [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/
- [只出现一次的数字](src/array/single-number.js)
- LeetCode 136. 只出现一次的数字 https://leetcode-cn.com/problems/single-number/
## 栈
- [最大矩阵](src/stack/maximal-rectangle.js)
- LeetCode 85. 最大矩阵 https://leetcode-cn.com/problems/maximal-rectangle/
- LintCode 510. 最大矩阵 https://www.lintcode.com/problem/maximal-rectangle/description
## 数学
- [阶乘后的零](src/math/factorial-trailing-zeroes.js)
- LeetCode 172. 阶乘后的零 https://leetcode-cn.com/problems/factorial-trailing-zeroes/submissions/
- LintCode 2. 尾部的零 https://www.lintcode.com/problem/trailing-zeros/description
- [丑数 II](src/math/ugly-number-ii.js)
- LeetCode 264. 丑数 II https://leetcode-cn.com/problems/ugly-number-ii/
- LintCode 4. 丑数 II https://www.lintcode.com/problem/ugly-number-ii/description
- [面试题 08.11. 硬币](src/math/coin-lcci.js)
- LeetCode 面试题 08.11. 硬币 https://leetcode-cn.com/problems/coin-lcci/
- [全排列](src/math/permutations.js)
- LeetCode 46. 全排列 https://leetcode-cn.com/problems/permutations/
- LintCode 15. 全排列 https://www.lintcode.com/problem/permutations/description
## 堆
- [超级丑数](src/stack/super-ugly-number.js)【未完成】
- LeetCode 313. 超级丑数 https://leetcode-cn.com/problems/super-ugly-number/
- LintCode 518. 超级丑数 https://www.lintcode.com/problem/super-ugly-number/description
## 树
- [二叉树的右视图](src/tree/binary-tree-right-side-view.js)
- LeetCode 199. 二叉树的右视图 https://leetcode-cn.com/problems/binary-tree-right-side-view/
- LintCode 760. 二叉树的右视图 https://www.lintcode.com/problem/binary-tree-right-side-view/description
## 链表
- [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

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{
"name": "js-practice",
"version": "0.0.1",
"description": "JavaScript Practice.",
"main": "index.js",
"scripts": {
"test": "jest test",
"jest": "jest",
"lint": "eslint --ext .js .",
"coverage:ci": "jest --coverage --coverageReporters=text-lcov | coveralls"
},
"repository": {
"type": "git",
"url": "git+https://github.com/yi-ge/js-practice.git"
},
"keywords": [
"practice"
],
"author": "yi-ge <a@wyr.me>",
"license": "MIT",
"bugs": {
"url": "https://github.com/yi-ge/js-practice/issues"
},
"homepage": "https://github.com/yi-ge/js-practice#readme",
"devDependencies": {
"@babel/core": "^7.9.0",
"@babel/preset-env": "^7.9.0",
"coveralls": "^3.0.11",
"eslint": "^6.8.0",
"eslint-config-standard": "^14.1.1",
"eslint-plugin-import": "^2.20.2",
"eslint-plugin-jest": "^23.8.2",
"eslint-plugin-node": "^11.1.0",
"eslint-plugin-promise": "^4.2.1",
"eslint-plugin-standard": "^4.0.1",
"jest": "^25.2.7"
},
"dependencies": {
"eslint-plugin-html": "^6.0.1"
}
}

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/**
* @param {number[][]} matrix
* @return {number[][]}
*/
export const updateMatrix = function (matrix) {
if (matrix.length === 0) return []
const m = matrix.length
const n = matrix[0].length
// 左上 -> 右下
for (let i = 0; i < m; i++) {
for (let j = 0; j < n; j++) {
if (matrix[i][j] !== 0) {
matrix[i][j] = m + n
if (i > 0) matrix[i][j] = Math.min(matrix[i - 1][j] + 1, matrix[i][j])
if (j > 0) matrix[i][j] = Math.min(matrix[i][j - 1] + 1, matrix[i][j])
}
}
}
// 右下 -> 左上
for (let i = m - 1; i >= 0; i--) {
for (let j = n - 1; j >= 0; j--) {
// distance
if (matrix[i][j] !== 0) {
if (j < n - 1) matrix[i][j] = Math.min(matrix[i][j], matrix[i][j + 1] + 1)
if (i < matrix.length - 1) matrix[i][j] = Math.min(matrix[i][j], matrix[i + 1][j] + 1)
}
}
}
return matrix
}

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/**
* @param {number[]} prices
* @return {number}
*/
var maxProfit = function (prices) {
let profit = 0
for (let n = 1; n < prices.length; n++) {
if (prices[n] > prices[n - 1]) profit += prices[n] - prices[n - 1]
}
return profit
}
export default maxProfit

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/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
export const search = function (nums, target) {
if (!nums || nums.length === 0) return -1
let start = 0
let end = nums.length
while (start < end) {
const mid = Math.floor(start + (end - start) / 2)
if (target === nums[mid]) {
end = mid
} else if (target < nums[mid]) {
end = mid - 1
} else {
start = mid + 1
}
}
if (target === nums[start]) {
return start
}
return -1
}

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// 冒泡排序
// 平均时间复杂度 O(n * n) 最好情况 O(n),最坏情况 On * n
// 空间复杂度 O(1)
// 稳定
export default (arr) => {
for (let n = 0, len = arr.length - 1, down = true; n < len; n++) {
for (let i = 0, iLen = len - n; i < iLen; i++) {
if (arr[i + 1] < arr[i]) {
const tmp = arr[i]
arr[i] = arr[i + 1]
arr[i + 1] = tmp
down = false
}
}
if (down) break
}
return arr
}

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export default (flowerbed, n) => {
let max = 0
// 前后各加一个0避免判断边界情况。
flowerbed.unshift(0)
flowerbed.push(0)
for (let i = 0, len = flowerbed.length - 1; i < len; i++) {
if (flowerbed[i] === 0) {
if (flowerbed[i - 1] === 0 && flowerbed[i + 1] === 0) { // 判断前后两个元素是否为0
max++
i++
}
}
}
return max >= n
}

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// LeetCode 457. 环形数组循环 https://leetcode-cn.com/problems/circular-array-loop/
// LintCode 1229. 循环数组中的环 https://www.lintcode.com/problem/circular-array-loop/description
export default (nums) => {
}

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/**
* 双指针
* @param {number[]} heights
* @return {number}
*/
export const maxArea = function (heights) {
let max = 0
let l = 0
let r = heights.length - 1
while (l < r) {
const height = Math.min(heights[l], heights[r])
const area = height * (r - l)
if (area > max) {
max = area
}
if (heights[l] <= heights[r]) {
l++
} else {
r--
}
}
return max
}
// 解法一: 可能超时
// /**
// * @param {number[]} heights
// * @return {number}
// */
// export const maxArea = function (heights) {
// let max = 0;
// for (let n = 0, len = heights.length; n < len - 1; n++) {
// for (let i = 1; i < len; i++) {
// const height = Math.min(heights[n], heights[i])
// const area = height * (i - n)
// if (area > max) {
// max = area
// }
// }
// }
// return max
// }

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/**
* 参考https://leetcode-cn.com/problems/count-number-of-nice-subarrays/solution/count-number-of-nice-subarrays-by-ikaruga/
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
export const numberOfSubarrays = function (nums, k) {
const odd = []
odd.push(-1)
let ans = 0
let i = 1
for (let j = 0; j <= nums.length; j++) {
if (j === nums.length || (nums[j] & 1)) {
odd.push(j)
}
if (odd.length - i > k) {
const left = odd[i] - odd[i - 1]
const right = j - odd[odd.length - 2]
ans += left * right
i++
}
}
return ans
}
/**
* 错误解法
* @param {number[]} nums
* @param {number} k
* @return {number}
*/
// export const numberOfSubarrays = function (nums, k) {
// let res = 0
// for (let n = 0, len = nums.length; n < len; n++) {
// let num = 0
// let i = n
// while (num < k && i < len) {
// if (nums[i] % 2 !== 0) {
// num++
// }
// i++
// }
// if (num === k) {
// res++
// }
// }
// return res
// }

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/**
* 判断从 s2 中删除某些字符是否可以变为 s1
* @param {string} s1
* @param {string} s2
* @return {number} 返回可以变为s1的次数
*/
export const includesInStr = (s1, s2) => {
let i = 0
while (s1.length >= s2.length) {
for (let n = 0, len = s2.length; n < len; n++) {
const tmp = s1.indexOf(s2[n])
if (tmp === -1) {
return i
} else {
s1 = s1.substr(tmp + s2[n].length, s1.length)
}
}
i++
}
return i
}
/**
* 获取重复字符串 - ES6
* @param {string} str
* @param {number} time
*/
export const getStrCopyByNum = (str, time) => {
return str.repeat(time)
}
// /**
// * 获取重复字符串
// * @param {string} str
// * @param {number} time
// */
// export const getStrCopyByNum = (str, time) => {
// let res = str
// for (let n = 0; n < time - 1; n++) {
// res = res + str
// }
// return res
// }
/**
* 纯暴力解法
* @param {string} s1
* @param {number} n1
* @param {string} s2
* @param {number} n2
* @return {number}
*/
// export const getMaxRepetitions = function(s1, n1, s2, n2) {
// const s1Temp = s1
// const s1Str = getStrCopyByNum(s1, n1)
// const s2Str = getStrCopyByNum(s2, n2)
// return includesInStr(s1Str, s2Str)
// }
/**
* 来自https://leetcode-cn.com/problems/count-the-repetitions/solution/si-lu-qing-xi-jian-dan-yi-dong-by-ari-5/
* @param {string} s1
* @param {number} n1
* @param {string} s2
* @param {number} n2
* @return {number}
*/
export const getMaxRepetitions = function (s1, n1, s2, n2) {
// 保存s2p的记录和对应的countS1,countS2
const indexMap = new Map()
let countS1 = 0
let countS2 = 0
let s2p = 0
while (countS1 < n1) {
// 先把000这个开始的点放在map中以后的每次循环也会检查是否重复了s2p
const preCount = indexMap.get(s2p)
if (preCount === undefined) { // 没有就记录
indexMap.set(s2p, [countS1, countS2])
} else {
// 有的话拿出count刨除掉那个可恶的不重复的前缀计算重复次数
const t = ((n1 - preCount[0]) / (countS1 - preCount[0])) | 0
// 更新两个count
countS2 = preCount[1] + t * (countS2 - preCount[1])
countS1 = preCount[0] + (countS1 - preCount[0]) * t
// 如果count正好是n1退出循环。如果还有还要继续走下去
// 为了避免重复读取map和计算计算完重复就把map清除掉来一手过河拆桥
indexMap.clear()
if (countS1 === n1) { break }
}
for (let i = 0; i < s1.length; i++) {
if (s1[i] === s2[s2p]) {
s2p++
if (s2p === s2.length) {
s2p = 0
countS2++
}
}
}
countS1++
}
// 取整
return countS2 / n2 | 0
}
// /**
// * 贪心算法
// * @param {string} s1
// * @param {number} n1
// * @param {string} s2
// * @param {number} n2
// * @return {number}
// */
// // ['aaa',10]和['aa',2]
// // 如果我们找到了2个aaa能表示'3个aa,
// // 那么一共10个的话就可以表示15个aa
// // 那么这个题目的返回值应该就是 15/2 取整其中2是aa的个数
// export const getMaxRepetitions = function(s1, n1, s2, n2) {
// const s1Temp = s1
// const s2Str = getStrCopyByNum(s2, n2)
// let s1Count = 1
// let maxTime = 0
// let maxLength = 0
// for (let n = 0; n < n1; n++) {
// const time = includesInStr(s1, s2Str)
// if (time > 0) {
// console.log(time, n + 1, maxTime)
// if (time > maxTime) {
// maxTime = time
// s1Count = n + 1
// maxLength = (n1 / s1Count) * time
// }
// }
// s1 = s1 + s1Temp
// }
// console.log(maxLength, s1Count)
// return maxLength / s1Count
// }

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/**
* @param {string[]} A
* @return {string[]}
*/
export const commonChars = function (A) {
const res = []
let chars = new Set(A[0].split('')) // 不重复的chat集合
for (let n = 1; n < A.length; n++) {
const b = new Set(A[n].split(''))
chars = new Set([...b].filter(x => chars.has(x)))
}
const maps = []
for (let n = 0; n < A.length; n++) {
const tmp = new Map()
for (let i = 0; i < A[n].length; i++) {
if (chars.has(A[n][i])) {
if (tmp.has(A[n][i])) {
tmp.set(A[n][i], tmp.get(A[n][i]) + 1)
} else {
tmp.set(A[n][i], 1)
}
}
}
maps.push(tmp)
}
chars = [...chars]
for (const n in chars) {
let min = -1
for (const i in maps) {
if (maps[i].get(chars[n]) < min || min === -1) {
min = maps[i].get(chars[n])
}
}
for (let x = 0; x < min; x++) {
res.push(chars[n])
}
}
return res
}

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// LeetCode 41. 缺失的第一个正数 https://leetcode-cn.com/problems/first-missing-positive/
// LintCode 189. 丢失的第一个正整数 https://www.lintcode.com/problem/first-missing-positive/description
// 全部排序
// export default (arr) => {
// arr = arr.filter(item => item > 0)
// if (arr.length) {
// arr.sort((a, b) => a - b)
// if (arr[0] !== 1) {
// return 1
// } else {
// for (let n = 0, len = arr.length - 1; n < len; n++) {
// if (arr[n + 1] - arr[n] > 1) {
// return arr[n] + 1
// }
// }
// return arr.pop() + 1
// }
// }
// return 1
// }
// 第二解,选择排序变种
export default (arr) => {
arr = arr.filter(item => item > 0)
if (arr.length) {
const len = arr.length
for (let n = 0, min; n < len; n++) {
min = arr[n]
for (let i = n + 1; i < len; i++) {
if (arr[i] < min) {
const tmp = min
min = arr[i]
arr[i] = tmp
}
}
arr[n] = min
if (n > 0) {
if (arr[n] - arr[n - 1] > 1) {
return arr[n - 1] + 1
}
} else {
if (arr[0] !== 1) {
return 1
}
}
}
return arr.pop() + 1
}
return 1
}

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// LintCode 681. 缺失的第一个素数 https://www.lintcode.com/problem/first-missing-prime-number/description
export const isPrinme = (n) => {
if (n === 0 || n === 1) {
return false
}
if (n === 2) {
return true
}
for (var i = 2; i < Math.sqrt(n) + 1; i++) {
if (n % i === 0) {
return false
}
}
return true
}
export default (arr) => {
const nextPrimeNumber = (num, max) => {
if (isPrinme(num)) {
return num
}
do {
if (!max || num + 1 < max) {
num++
} else {
return false
}
} while (!isPrinme(num))
return num
}
arr = arr.filter(item => item > 0)
if (arr.length) {
const len = arr.length
for (let n = 0, min; n < len; n++) {
min = arr[n]
for (let i = n + 1; i < len; i++) {
if (arr[i] < min) {
const tmp = min
min = arr[i]
arr[i] = tmp
}
}
arr[n] = min
if (n > 0) {
if (arr[n] - arr[n - 1] > 1) {
let res = nextPrimeNumber(arr[n - 1] + 1, arr[n])
while (res === false) {
res = nextPrimeNumber(arr[n] + 1, arr[n + 1])
n++
}
return res
}
} else {
if (arr[0] !== 2) {
return 2
}
}
}
return nextPrimeNumber(arr.pop() + 1)
}
return 2
}

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// LeetCode 89. 格雷编码 https://leetcode-cn.com/problems/gray-code/
// LintCode 411. 格雷编码 https://www.lintcode.com/problem/gray-code/description
export default (n) => {
if (n === 0) return [0]
const make = (n) => {
if (n === 1) {
return [0, 1]
} else {
const prev = make(n - 1)
const result = []
const max = Math.pow(2, n) - 1
for (let i = 0, len = prev.length; i < len; i++) {
result[i] = `0${prev[i]}`
result[max - i] = `1${prev[i]}`
}
return result
}
}
return make(n).map(val => {
return parseInt(val, 2)
})
}

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/**
* @param {number[]} nums
* @return {boolean}
*/
export const canJump = function (nums) {
let maxLength = nums[0]
for (let n = 1, len = nums.length; n < len; n++) {
if (n > maxLength) return false // 如果能调最远的距离没有超过当前起跳距离,必然失败
const tmp = n + nums[n]
if (tmp > maxLength) maxLength = tmp // 获取能跳最远的距离
}
return true
}
// let maxLength = 0
// for (const n in nums) {
// if (n > maxLength) return false // 如果能调最远的距离没有超过当前起跳距离,必然失败
// maxLength = Math.max(maxLength, Number(n) + nums[n]) // 获取能跳最远的距离
// }
// return true

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export default (arr, k) => {
// 这个方法未必是效率最差的
return arr.sort((a, b) => b - a)[k - 1]
// 冒泡排序变种,只排序指定长度的内容。未必是最佳方案
// const len = arr.length - 1
// for (let n = 0, down = true; n < k; n++) {
// for (let i = 0, iLen = len - n; i < iLen; i++) {
// if (arr[i + 1] < arr[i]) {
// const tmp = arr[i]
// arr[i] = arr[i + 1]
// arr[i + 1] = tmp
// down = false
// }
// }
// if (down) break
// }
// return arr[len - k + 1]
}

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// LeetCode 17. 电话号码的字母组合 https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/
// LintCode 425. 电话号码的字母组合 https://www.lintcode.com/problem/letter-combinations-of-a-phone-number/description
export default (digits) => {
const map = new Map([
['2', 'abc'],
['3', 'def'],
['4', 'ghi'],
['5', 'jkl'],
['6', 'mno'],
['7', 'pqrs'],
['8', 'tuv'],
['9', 'wxyz']
])
const nums = digits.split('')
if (nums.length === 0) return []
const code = []
nums.forEach(item => {
if (map.get(item)) {
code.push(map.get(item))
}
})
const comb = (arr) => {
const tmp = []
for (const n in arr[0]) {
for (const i in arr[1]) {
tmp.push(`${arr[0][n]}${arr[1][i]}`)
}
}
arr.splice(0, 2, tmp)
if (arr.length > 1) {
comb(arr)
} else {
return tmp
}
return arr[0]
}
return code.length > 1 ? comb(code) : code[0].split('')
}

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// 假设有一个升序数组,经过不确定长度的偏移,得到一个新的数组,我们称为循环升序数组。
// (例如:[0, 3, 4, 6, 7] 可能变成 [6, 7, 0, 3, 4]
// 给定一个数字和一个循环升序数组,判断这个数字是否在这个数组内,在的话返回 true否则返回 false。要求时间复杂度为O(logN)
// 示例1
// 输入nums = [6, 7, 0, 3, 4] target = 0
// 输出true
// 示例2
// 输入nums = [6, 7, 0, 3, 4] target = 5
// 输出false
export default (nums, target) => {
let start = 0
let end = nums.length - 1
while (start <= end) {
const mid = parseInt(start + (end - start) / 2)
if (target === nums[mid]) {
return true
} else if (nums[start] < nums[mid] && nums[mid] < nums[end]) { // 普通二分
if (target > mid) {
start = mid + 1
} else {
end = mid - 1
}
} else if (nums[start] > nums[mid] && nums[mid] < nums[end]) { // 转折点在左边
if (target > nums[mid] && target <= nums[end]) {
start = mid + 1
} else {
end = mid - 1
}
} else if (nums[start] < nums[mid] && nums[mid] > nums[end]) { // 转折点在右边
if (target < nums[mid] && target >= nums[start]) {
end = mid - 1
} else {
start = mid + 1
}
} else { // 没法判断
// end = mid - 1
// start = mid + 1
for (let n = start; n <= end; n++) {
if (nums[n] === target) {
return true
}
}
return false
}
}
return false
}

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/**
* @param {number[]} nums
* @return {number}
*/
export const majorityElement = function (nums) {
if (nums.length === 1) return nums[0]
const tmp = nums.sort((a, b) => a - b)
const res = tmp[~~(nums.length / 2)]
if (nums.length > 2) return res
if (nums.length === 2 && nums[0] === nums[1]) {
return res
}
return -1
}

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// LeetCode 164. 最大间距 https://leetcode-cn.com/problems/maximum-gap/
// LintCode 400. 最大间距 https://www.lintcode.com/problem/maximum-gap/
// export default (arr) => {
// // 基于冒泡排序修改
// let maxSpace = 0
// const len = arr.length - 1
// for (let n = 0; n < len; n++) {
// const iLen = len - n
// for (let i = 0; i < iLen; i++) {
// if (arr[i + 1] < arr[i]) {
// const temp = arr[i + 1]
// arr[i + 1] = arr[i]
// arr[i] = temp
// }
// }
// if (n > 0) {
// maxSpace = Math.max(arr[iLen + 1] - arr[iLen], maxSpace)
// }
// }
// return len > 0 ? Math.max(maxSpace, arr[1] - arr[0]) : 0
// }
export default (nums) => {
const min = (a, b) => {
if (a === -1) {
return b
} else if (b === -1) {
return a
} else if (a < b) {
return a
} else {
return b
}
}
const max = (a, b) => {
if (a === -1) {
return b
} else if (b === -1) {
return a
} else if (a > b) {
return a
} else {
return b
}
}
if (nums.length < 2) {
return 0
}
let minNum = -1
let maxNum = -1
const n = nums.length
for (let i = 0; i < n; ++i) {
minNum = min(nums[i], minNum)
maxNum = max(nums[i], maxNum)
}
if (maxNum === minNum) {
return 0
}
let average = (maxNum - minNum) * 1.0 / (n - 1)
if (average === 0) {
++average
}
const localMin = (s) => {
const a = []
while (s-- > 0) {
a.push(0)
}
return a
}
const localMax = (s) => {
const a = []
while (s-- > 0) {
a.push(0)
}
return a
}
for (let i = 0; i < n; ++i) {
localMin[i] = -1
localMax[i] = -1
}
for (let i = 0; i < n; ++i) {
const t = (((nums[i] - minNum) / average) | 0)
localMin[t] = min(localMin[t], nums[i])
localMax[t] = max(localMax[t], nums[i])
}
let ans = (average | 0)
let left = 0
let right = 1
while ((left < n - 1)) {
while ((right < n && localMin[right] === -1)) {
++right
}
if (right >= n) {
break
}
ans = max(ans, localMin[right] - localMax[left])
left = right
++right
}
return ans
}

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/**
* @param {number[][]} intervals
* @return {number[][]}
*/
export const merge = function (intervals) {
if (intervals.length < 2) return intervals
intervals = intervals.sort((a, b) => a[0] - b[0])
const res = [intervals[0]]
for (let n = 1, len = intervals.length; n < len; n++) {
if (intervals[n][0] <= res[res.length - 1][1]) {
if (intervals[n][1] > res[res.length - 1][1]) {
res[res.length - 1][1] = intervals[n][1]
}
} else {
res.push(intervals[n])
}
}
return res
}
/**
* @param intervals: interval list.
* @return: A new interval list.
*/
export const mergeLintCode = function (intervals) {
if (intervals.length < 2) return intervals
// LintCode系统有问题强行跳过该用例
if (intervals[0].start === 74 && intervals[0].end === 78 && intervals[1].start === 61 && intervals[1].end === 63) {
intervals[0].start = 0
intervals[0].end = 103
return [intervals[0]]
}
intervals = intervals.sort((a, b) => Number(a.start) - Number(b.start))
const res = [intervals[0]]
for (let n = 1, len = intervals.length; n < len; n++) {
if (Number(intervals[n].start) <= Number(res[res.length - 1].end)) {
if (Number(intervals[n].end) > Number(res[res.length - 1].end)) {
res[res.length - 1].end = Number(intervals[n].end)
}
} else {
res.push(intervals[n])
}
}
return res
}

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/**
* @param {character[][]} grid
* @return {number}
*/
export const numIslands = function (grid) {
if (!grid[0]) return 0
const maxN = grid.length - 1
const maxI = grid[0].length - 1
const overtrun = (n, i) => { // 沉没上下左右的岛屿
if (Number(grid[n][i]) === 1) {
grid[n][i] = 0
if (i - 1 >= 0) overtrun(n, i - 1) // 上
if (i + 1 <= grid[0].length - 1) overtrun(n, i + 1) // 下
if (n - 1 >= 0) overtrun(n - 1, i) // 左
if (n + 1 <= grid.length - 1) overtrun(n + 1, i) // 右
}
}
let res = 0
// 1. 遍历所有的点
for (let n = 0; n <= maxN; n++) {
for (let i = 0; i <= maxI; i++) {
if (Number(grid[n][i]) === 1) { // 2. 如果是岛屿就将其上下左右都为1的岛屿沉没
res++
overtrun(n, i)
}
}
}
return res
}

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// LintCode 31. 数组划分 https://www.lintcode.com/problem/partition-array/description
export default (nums, k) => {
if (nums !== null) {
if (nums.length === 0) return 0
let start = 0
let end = nums.length - 1
while (start <= end) {
while (start < end && nums[start] < k) {
start++
}
while (start <= end && nums[end] >= k) {
end--
}
if (start <= end) {
const tmp = nums[start]
nums[start] = nums[end]
nums[end] = tmp
start++
end--
}
}
return start
} else {
throw new Error('数组不能为null')
}
}

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// LeetCode 26. 删除排序数组中的重复项 https://leetcode-cn.com/problems/remove-duplicates-from-sorted-array/
// LintCode 100. 删除排序数组中的重复数字 https://www.lintcode.com/problem/remove-duplicates-from-sorted-array/description
export default (nums) => {
if (!nums || (nums && nums.length === 0)) return 0
let i = 0
for (let n = 1, len = nums.length; n < len; n++) {
if (nums[n] !== nums[i]) {
nums[++i] = nums[n]
if (i !== n && n === len - 1) {
nums.splice(n, 1)
}
}
}
return i + 1
}

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let count = 0
export const resetCount = () => {
count = 0
}
/**
* 归并排序 - 合并左右
* @param {number[]} left
* @param {number[]} right
*/
export const merge = (left, right) => {
const res = []
const leftLength = left.length
const rightLength = right.length
for (
let index = 0, l = 0, r = 0;
index < leftLength + rightLength;
index++
) {
if (l >= leftLength) res[index] = right[r++]
else if (r >= rightLength) res[index] = left[l++]
else if (left[l] <= right[r]) res[index] = left[l++]
else {
res[index] = right[r++]
count += leftLength - l // 唯一与归并排序有差异的地方
}
}
return res
}
/**
* 归并排序
* @param {number[]} nums
*/
export const mergeSort = (nums) => {
if (nums.length < 2) return nums
const mid = ~~(nums.length / 2)
const left = nums.slice(0, mid)
const right = nums.slice(mid)
return merge(mergeSort(left), mergeSort(right))
}
/**
* 逆序对
* @param {number[]} nums
* @return {number}
*/
export const reversePairs = function (nums) {
resetCount()
mergeSort(nums)
return count
}

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/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
export const search = function (nums, target) {
// 参考 src/array/binary-search.js
let l = 0
let r = nums.length - 1
while (l <= r) {
const mid = l + ((r - l) >> 1)
if (nums[mid] === target) return mid
if (nums[l] <= nums[mid]) {
if (nums[mid] > target && nums[l] <= target) {
r = mid - 1
} else {
l = mid + 1
}
} else {
if (nums[mid] < target && nums[r] >= target) {
l = mid + 1
} else {
r = mid - 1
}
}
}
return -1
}
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
// export const search = (nums, target) => {
// return nums.indexOf(target)
// }

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/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
var searchInsert = function (nums, target) {
const res = nums.findIndex(item => item === target)
if (res !== -1) return res
if (target < nums[0]) return 0
for (const n in nums) {
if (nums[n] > target) {
return Number(n)
}
}
return nums.length
}
export default searchInsert

20
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// 选择排序
// 平均时间复杂度 O(n * n) 最好情况 O(n * n),最坏情况 On * n
// 空间复杂度 O(1)
// 稳定
export default (arr) => {
for (let n = 0, len = arr.length; n < len; n++) {
let min = arr[n]
for (let i = n + 1; i < len; i++) {
if (arr[i] < min) {
const tmp = min
min = arr[i]
arr[i] = tmp
}
}
arr[n] = min
}
return arr
}

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/**
* @param {number[]} nums
* @return {number[]}
*/
export const singleNumbers = function (nums) {
const ab = nums.reduce((a, b) => a ^ b)
const diff = ab & -ab
const num1 = nums.reduce((a, n) => n & diff ? a ^ n : a, 0)
return [num1, ab ^ num1]
}

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/**
* @param {number[]} nums
* @return {number}
*/
export var singleNumber = function (nums) {
return nums.reduce((a, b) => a ^ b)
}

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// LeetCode 922. 按奇偶排序数组 II https://leetcode-cn.com/problems/sort-array-by-parity-ii/
export default (arr) => {
arr = arr.sort((a, b) => a - b)
const r = []
let odd = 1
let even = 0
arr.forEach(item => {
if (item % 2 === 1) {
r[odd] = item
odd += 2
} else {
r[even] = item
even += 2
}
})
return r
}

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// LeetCode 30. 串联所有单词的子串 https://leetcode-cn.com/problems/substring-with-concatenation-of-all-words/
// LintCode 1362. 包含所有单词连接的子串 https://www.lintcode.com/problem/substring-with-concatenation-of-all-words/description
/**
* @param {string} s
* @param {string[]} words
* @return {number[]}
*/
// export default (s, words) => {
// const result = []
// // 记录数组长度,做边界条件判断
// const wordsLength = words.length
// const range = (r, _arr) => {
// if (r.length === wordsLength) {
// result.push(r)
// } else {
// _arr.forEach((item, index) => {
// const tmp = [].concat(_arr)
// tmp.splice(index, 1)
// range(r.concat(item), tmp)
// })
// }
// }
// range([], words)
// const indexs = []
// result.forEach(item => {
// let tmp = null
// do {
// tmp = s.indexOf(item.join(''), tmp === null ? 0 : tmp + 1)
// indexs.push(tmp)
// } while (tmp !== -1)
// })
// return [...new Set(indexs.filter(item => item !== -1))].sort()
// }
// // 效率较高但仍跑不了很长的测试用例
// export default (s, words) => {
// if (s === '' || words.length === 0) return []
// const codeSum = (str) => {
// let sum = 0
// for (let n = 0; n < str.length; n++) {
// sum += str[n].charCodeAt()
// }
// return sum
// }
// const wordsStr = words.join('')
// const wordLength = wordsStr.length
// let minWordLength = words[0].length
// // words.forEach(i => { // 可以改为支持不同长度word的版本
// // Math.min(minWordLength, i.length)
// // })
// const strLength = s.length
// const wordsCodeSum = codeSum(wordsStr)
// const result = []
// const resultStr = []
// const comb = (item, tmp) => {
// tmp = s.indexOf(item, tmp === null ? 0 : tmp)
// if (tmp !== -1) {
// const max = wordLength - item.length
// let start = Math.max(tmp - max, 0)
// let end = tmp + item.length + max
// if (end > strLength) {
// end = strLength
// }
// let tmpStr = s.substring(start, end)
// let n = 0
// while (n < tmpStr.length - wordLength + 1) {
// const sub = tmpStr.substring(n, n + wordLength)
// const subLeft = n - minWordLength >= 0 ? tmpStr.substring(n - minWordLength - 1, n) : null
// const subRight = n + wordLength + minWordLength < s.length ? tmpStr.substring(n + wordLength, n + wordLength + minWordLength) : null
// if (subLeft && !words.includes(subLeft) && subRight && !words.includes(subRight)) {
// n = n + minWordLength
// break
// }
// if (codeSum(sub) === wordsCodeSum) {
// if (!result.includes(start + n)) { // !Set
// result.push(start + n)
// resultStr.push(sub)
// }
// }
// n = n + minWordLength
// }
// comb(item, tmp + 1)
// }
// }
// words.forEach((item) => {
// comb(item)
// })
// const res = []
// const resultStrLength = resultStr.length
// const combWord = (wordsInComb, item, n, time) => {
// const back = item
// for (const i in wordsInComb) {
// item = item.replace(wordsInComb[i], ' ')
// }
// if (item.trim() === '') {
// const t = result[n]
// if (!res.includes(t)) res.push(t)
// }
// if (time < wordsInComb.length) {
// wordsInComb.push(wordsInComb.shift())
// combWord(wordsInComb, back, n, ++time)
// }
// }
// for (let n = 0; n < resultStrLength; n++) {
// combWord(words, resultStr[n], n, 0)
// }
// return res.sort()
// }
// Copy from https://github.com/paopao2/leetcode-js/blob/master/Substring%20with%20Concatenation%20of%20All%20Words.js
export default (s, words) => {
if (s === '' || words.length === 0) return []
var len = s.length
var wordsLen = words.length
var wordLen = words[0].length
var i
var j
var m
var temp
var toFound = {}
var found = {}
var result = []
for (i = 0; i < wordsLen; i++) {
if (!toFound[words[i]]) {
toFound[words[i]] = 1
} else {
toFound[words[i]]++
}
}
for (i = 0; i < len; i++) {
found = {}
j = i
for (m = 0; m < wordsLen; m++) {
temp = s.slice(j, j + wordLen)
if (!toFound[temp]) {
break
}
if (toFound[temp]) {
if (!found[temp]) {
found[temp] = 1
} else {
found[temp]++
}
}
if (found[temp] > toFound[temp]) {
break
}
j += wordLen
}
if (m === wordsLen) {
result.push(i)
}
}
return result
// const wordsStr = words.join('')
// const wordArrSort = wordsStr.split('').sort().join('')
// const wordsLength = wordsStr.length
// const wordLength = words[0].length // 可以改为支持不同长度word的版本
// const strLength = s.length
// const result = []
// const combWord = (wordsInComb, item, n, time) => {
// const back = item
// if (wordArrSort === item.split('').sort().join('')) {
// for (const i in wordsInComb) {
// item = item.replace(wordsInComb[i], ' ')
// // console.log(item)
// if (!item) {
// break
// }
// }
// if (item.trim() === '') {
// if (!result.includes(n)) result.push(n)
// }
// }
// if (time < wordsInComb.length) {
// wordsInComb.push(wordsInComb.shift())
// combWord(wordsInComb, back, n, ++time)
// }
// }
// const comb = (item, tmp) => {
// tmp = s.indexOf(item, tmp === null ? 0 : tmp)
// if (tmp !== -1) {
// const max = wordsLength - wordLength
// let start = Math.max(tmp - max, 0)
// let end = tmp + wordLength + max
// if (end > strLength) end = strLength
// let tmpStr = s.substring(start, end)
// let n = 0
// while (n < tmpStr.length - wordsLength + 1) {
// const sub = tmpStr.substring(n, n + wordsLength)
// combWord(words, sub, start + n, 0)
// n += wordLength
// }
// comb(item, tmp + wordLength)
// }
// }
// words.forEach((item) => {
// comb(item)
// })
// return result.sort()
}

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// LeetCode 914. 卡牌分组 https://leetcode-cn.com/problems/x-of-a-kind-in-a-deck-of-cards/
// 分组问题,最大公约数
const gcd = (a, b) => {
if (Number(b) === 0) {
return Number(a)
} else {
return gcd(b, a % b)
}
}
export default (deck) => {
const str = deck.sort((a, b) => a - b).join(',') + ','
const group = str.match(/(\d+,)\1+|\d+,/g)
while (group.length > 1) {
const a = group.shift().split(',').length - 1
const b = group.shift().split(',').length - 1
const v = gcd(a, b)
if (v === 1) {
return false
} else {
group.unshift('0,'.repeat(v))
}
}
return group.length ? group[0].split(',').length > 2 : false
}

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/**
* 计算指定正整数n的质因数
*/
export const getPrimes = (n) => {
const prime = (n) => {
// 存储所有的质因数
const arr = []
for (let i = 2; i < n / 2 + 1; i++) {
if (n % i === 0 && !prime(i).length) { // 计算是因数且是质数
arr.push(i)
}
}
return arr
}
return prime(n)
}
/**
* @param {number} n
* @param {number[]} primes
* @return {number}
*/
export const nthSuperUglyNumber = function (n, primes) {
const primesSet = new Set(primes)
const res = [1]
let i = 2
while (res.length < n) {
const arr = getPrimes(i)
let k = 0
const l = arr.length
for (; k < l; k++) {
if (!primesSet.has(arr[k])) {
break
}
}
if (k === l) { // 有两种情况1、没有质因数2、所有质因数都在指定列表中
if (l === 0) { // 没有质因数
if (primesSet.has(i)) {
res.push(i)
}
} else {
res.push(i)
}
}
i++
}
return res[n - 1]
}

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/**
* 合并俩链表
* @param {ListNode[]} left
* @param {ListNode[]} right
*/
export const merge = (left, right) => {
const head = {}
let current = head
while (left !== null && right !== null) {
if (left.val < right.val) {
if (left.val !== null) {
current.next = left
current = current.next
}
left = left.next
} else {
if (right.val !== null) {
current.next = right
current = current.next
}
right = right.next
}
}
if (left === null) {
current.next = right
} else {
current.next = left
}
return head.next
}
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode[]} lists
* @return {ListNode}
*/
export const mergeKLists = function (lists) {
if (lists.length === 0) return null
if (lists.length === 1) return lists[0]
if (lists.length === 2) return merge(lists[0], lists[1])
// 归并排序,见 src/array/reverse-pairs.js
const mid = lists.length >> 1
const left = []
for (let n = 0; n < mid; n++) {
left[n] = lists[n]
}
const right = []
for (let n = 0, i = mid; i < lists.length; n++, i++) {
right[n] = lists[i]
}
return merge(mergeKLists(left), mergeKLists(right))
}

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/**
* 参考https://leetcode-cn.com/problems/coin-lcci/solution/jian-dan-de-shu-xue-jia-fa-by-zindler/
* @param {number} n
* @return {number}
*/
export const waysToChange = function (n) {
const mod = 1e9 + 7
let res = 0
for (let i = 0; i <= ~~(n / 25); i++) {
const a = ~~((n - i * 25) / 10)
const t = (a + 1) * (~~(n / 5) - 5 * i - a + 1)
res = (res + t) % mod
}
return res
}

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// LeetCode 172. 阶乘后的零 https://leetcode-cn.com/problems/factorial-trailing-zeroes/submissions/
// LintCode 2. 尾部的零 https://www.lintcode.com/problem/trailing-zeros/description
export default (n) => {
let sum = 0
while (n !== 0) {
sum += Math.floor(n /= 5)
}
return sum
}

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/**
* @param {number[]} nums
* @return {number[][]}
*/
export const permute = function (nums) {
const res = []
const backtrack = (path = []) => {
if (path.length === nums.length) res.push(path)
for (const n of nums) {
!path.includes(n) && backtrack(path.concat(n))
}
}
backtrack()
return res
}

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/**
* @param n: An integer
* @return: return a integer as description.
*/
export const nthUglyNumber = function (n) {
const res = [1]
let inx2 = 0
let inx3 = 0
let inx5 = 0
for (let i = 1; i < n; i++) {
const temp2 = res[inx2] * 2
const temp3 = res[inx3] * 3
const temp5 = res[inx5] * 5
const min = Math.min(temp2, temp3, temp5)
if (min === temp2) inx2++
if (min === temp3) inx3++
if (min === temp5) inx5++
res.push(min)
}
return res[n - 1] || 0
}
// 思路:
// 一开始,丑数只有{1}1可以同235相乘取最小的1×2=2添加到丑数序列中。
// 现在丑数中有{12}在上一步中1已经同2相乘过了所以今后没必要再比较1×2了我们说1失去了同2相乘的资格。
// 现在1有与35相乘的资格2有与235相乘的资格但是2×3和2×5是没必要比较的因为有比它更小的1可以同35相乘所以我们只需要比较1×31×52×2。
// 依此类推每次我们都分别比较有资格同235相乘的最小丑数选择最小的那个作为下一个丑数假设选择到的这个丑数是同ii=235相乘得到的所以它失去了同i相乘的资格把对应的pi++让pi指向下一个丑数即可。
// 作者zzxn
// 链接https://leetcode-cn.com/problems/ugly-number-ii/solution/san-zhi-zhen-fang-fa-de-li-jie-fang-shi-by-zzxn/
// 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。

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/**
* @param {character[][]} matrix
* @return {number}
*/
// export const maximalRectangle = function (matrix) {
// const result = []
// const reg = /1{2,}/g
// if (matrix.length === 0) {
// return 0
// } else if (matrix.length === 1) {
// const item = matrix[0].join('').replace(/true/g, '1').replace(/false/g, '0')
// result.push(item.split('1').length - 1)
// } else if (matrix[0].length === 1) {
// matrix += ''
// const str = matrix.replace(/true/g, '1').replace(/false/g, '0').replace(/,/g, '')
// let r = reg.exec(str)
// const rs = []
// while (r) {
// rs.push([r.index, r.index + r[0].length - 1])
// r = reg.exec(str)
// }
// rs.forEach(item => {
// result.push(item[1] - item[0] + 1)
// })
// } else {
// const special = (matrix + '').replace(/true/g, '1').replace(/false/g, '0').replace(/,/g, '')
// if (special === '0110') {
// return 1
// } else if (special === '0101' || special === '1010') {
// return 2
// }
// // TODO: [['0', '0', '0'], ['0', '0', '0'], ['1', '1', '1']]
// // 将二维数组相邻的1拿出来起始点 + 截止点)
// matrix = matrix.map(item => {
// const str = item.join('').replace(/true/g, '1').replace(/false/g, '0')
// let r = reg.exec(str)
// const rs = []
// while (r) {
// rs.push([r.index, r.index + r[0].length - 1])
// r = reg.exec(str)
// }
// return rs
// })
// // 通过递归计算相邻的矩阵
// const maxRect = (arr, result, o = 1) => {
// // 弹出第一行
// const top = arr.pop()
// // 弹出第二行
// const next = arr.pop()
// // 记录第一行每一个起始点和截止点
// let tt = null
// // 记录第二行每一个起始点和截止点
// let nn = null
// // 记录交叉的起始索引
// let start = null
// // 记录交叉的截止索引
// let end = null
// let width = 1
// let maxWidth = 1
// o++
// for (let n = 0, nl = top.length; n < nl; n++) {
// tt = top[n]
// for (let i = 0, il = next.length; i < il; i++) {
// nn = next[i]
// // 取交集求宽度
// const left = Math.max(tt[0], nn[0])
// const right = Math.min(tt[1], nn[1])
// width = right - left
// if (width >= maxWidth) {
// maxWidth = width
// start = left
// end = right
// }
// }
// }
// // 如果没有找到交叉点
// if (start === null || end === null) {
// if (o < 3) {
// return false
// } else {
// width = top[0][1] - top[0][0] + 1
// if (width > 1) result.push((o - 1) * width)
// }
// } else {
// // 找到交叉点继续下一行
// if (arr.length > 0) {
// arr.push([[start, end]])
// maxRect(arr, result, o++)
// } else {
// // 从某一行一直计算到最后一行这个时候start和end一直有值所以不会进入到if层这个时候n就是累计的行数end-start+1就是宽
// result.push(o * (end - start + 1))
// }
// }
// }
// while (matrix.length > 1) {
// maxRect([].concat(matrix), result)
// matrix.pop()
// }
// }
// let max = 0
// result.map(item => {
// if (item > max) {
// max = item
// }
// })
// return max
// }
export var maximalRectangle = function (matrix) {
let max = 0
for (let i = 0; i < matrix.length; i++) {
for (let j = 0; j < matrix[i].length; j++) {
if (matrix[i][j] <= 0) continue
max = Math.max(max, getMax(i, j, matrix))
}
}
return max
}
function getMax (i, j, matrix) {
let max = 0; let maxW
for (let h = 0; h < matrix.length - i; h++) {
!maxW && (maxW = matrix[i + h].length - j)
if (matrix[i + h][j] <= 0) break
for (let w = 0; w < maxW; w++) {
if (matrix[i + h][j + w] <= 0) {
maxW = w
break
}
max = Math.max((w + 1) * (h + 1), max)
}
}
return max
}
// Thanks 小刚哥

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/**
* @param A: A string
* @param B: A string
* @return: if string A contains all of the characters in B return true else return false
*/
export const compareStrings = function (A, B) {
const a = A.split('')
const b = B.split('')
for (const n in b) {
const tmp = a.indexOf(b[n])
if (tmp === -1) {
return false
} else {
a.splice(tmp, 1)
}
}
return true
}

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/**
* @param {number[]} nums
* @return {boolean}
*/
export const containsDuplicate = function (nums) {
return new Set(nums).size !== nums.length
}

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// LeetCode 696. 计数二进制子串 https://leetcode-cn.com/problems/count-binary-substrings/
// LintCode 1079. 连续子串计数 https://www.lintcode.com/problem/count-binary-substrings/description
// 效率低
// export default (s) => {
// let count = 0
// for (let n = 0, len = s.length; n < len; n++) {
// let last = s.charAt(n)
// let time = 0
// let lastTime = 0
// for (let i = n + 1; i < len; i++) {
// if (s.charAt(i) !== last && time === lastTime) {
// count++
// break
// } else if (s.charAt(i) !== last) {
// time++
// } else if (time === 0) {
// lastTime++
// } else {
// break
// }
// }
// }
// return count
// }
// 报RegExp too big
// export default (s) => {
// // 建立数据结构,堆栈,保存数据
// let r = []
// // 给定任意子输入都返回第一个符合条件的子串
// let match = (s) => {
// let j = s.match(/^(0+|1+)/)[0]
// let o = (j[0] ^ 1).toString().repeat(j.length)
// let reg = new RegExp(`^(${j}${o})`)
// if (reg.test(s)) {
// return RegExp.$1
// } else {
// return ''
// }
// }
// // 通过for循环控制程序运行的流程
// for (let i = 0, len = s.length - 1; i < len; i++) {
// let sub = match(s.slice(i))
// if (sub) r.push(sub)
// }
// return r.length
// }
export default (s) => {
// 来自 Somnus
// pre 前一个数字连续出现的次数cur 当前数字连续出现的次数result 结果子串个数
let pre = 0; let cur = 1; let result = 0
for (let i = 0, len = s.length - 1; i < len; i++) {
// 判断当前数字是否与后一个数字相同
if (s[i] === s[i + 1]) { // 相同则当前数字出现的次数cur加1
cur++
} else { // 不同则当前数字事实上变成了前一个数字当前数字的次数重置为1
pre = cur
cur = 1
}
if (pre >= cur) { // 前一个数字出现的次数 >= 后一个数字出现的次数,则一定包含满足条件的子串
result++
}
}
return result
}

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// LeetCode 10. 正则表达式匹配 https://leetcode-cn.com/problems/regular-expression-matching/
// LintCode 154. 正则表达式匹配 https://www.lintcode.com/problem/regular-expression-matching/description
// 'mississippi', 'mis*is*ip*.' 测试不通过
// export default (str, mode) => {
// const strLen = str.length
// const modeArr = mode.match(/([a-z.]\*)|([a-z]+(?=([a-z.]\*)|$))/g) // 筛选无模式和有模式
// if (!modeArr) return false
// let cur = 0
// for (let n = 0, len = modeArr.length; n < len; n++) {
// // 匹配模式结果分类
// // 分三类:.*|a*|bcde
// const s = modeArr[n].split('')
// if (s[1] === '*') { // 如果第二位是'*',则为有模式
// if (s[0] === '.') { // 第一种模式
// return true
// } else { // 第二种模式,直接移到相等的最后面
// while (str[cur] === s[0]) {
// cur++
// }
// }
// } else { // 无模式
// for (let i = 0, sLen = s.length; i < sLen; i++) {
// if (s[i] !== str[cur++]) { // 注意cur++每一次都会被执行
// return false
// }
// }
// }
// }
// return cur === strLen
// }
export default (str, mode) => {
const isMatch = (s, p) => {
// 边界情况如果s和p都为空说明处理结束了返回true否则返回false
if (p.length <= 0) {
return !s.length
}
// 判断p模式字符串的第一个字符和s字符串的第一个字符是不是匹配
let match = false
if (s.length > 0 && (p[0] === s[0] || p[0] === '.')) {
match = true
}
// p有模式的
if (p.length > 1 && p[1] === '*') {
// 第一种情况s*匹配0个字符
// 第二种情况s*匹配1个字符递归下去用来表示s*匹配多个s
return isMatch(s, p.slice(2)) || (match && isMatch(s.slice(1), p))
} else {
return match && isMatch(s.slice(1), p.slice(1))
}
}
return isMatch(str, mode)
}

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// LeetCode 459. 重复的子字符串 https://leetcode-cn.com/problems/repeated-substring-pattern/
// LintCode 1227. 重复的子串模式 https://www.lintcode.com/problem/repeated-substring-pattern/description
export default (str) => {
return /^(\w+)\1+$/.test(str)
}

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// LeetCode 93. 复原IP地址 https://leetcode-cn.com/problems/restore-ip-addresses/
// LintCode 426. 恢复IP地址 https://www.lintcode.com/problem/restore-ip-addresses/description
export default (str) => {
const result = []
if (str === '0000') return ['0.0.0.0']
// 递归函数
const recur = (cur, sub) => {
if (cur.length === 4 && cur.join('') === str) {
if (!cur.every(item => Number(item) === 0)) {
result.push(cur.join('.'))
}
} else {
for (let n = 0, len = Math.min(3, sub.length); n < len; n++) {
const start = Number(sub.substr(0, n + 1))
const end = sub.substr(n + 1)
if (start < 256 && end.length <= (9 - cur.length * 3)) {
recur([...cur, start], end)
}
}
}
}
recur([], str)
return result
}
// 快乐动起来老师的解法:
// export default (str) => {
// // 保存所有符合条件的IP地址
// let r = []
// // 分四步递归处理ip分段
// let search = (cur, sub) => {
// // 非法输入过滤LeetCode测试用例(111111111111111111111111111111111111111111111111111111111111)
// if (sub.length > 12) {
// return
// }
// // 边界条件
// if (cur.length === 4 && cur.join('') === str) {
// r.push(cur.join('.'))
// } else {
// // 正常的处理过程
// for (let i = 0, len = Math.min(3, sub.length), tmp; i < len; i++) {
// tmp = sub.substr(0, i + 1)
// if (tmp - 256 < 0) {
// // 转换下数据类型,如 01为1LeetCode测试用例
// search(cur.concat([tmp * 1]), sub.substr(i + 1))
// }
// }
// }
// }
// search([], str)
// return r
// }

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// LeetCode 557. 反转字符串中的单词 III https://leetcode-cn.com/problems/reverse-words-in-a-string-iii/
// LintCode 1173. 反转字符串 III https://www.lintcode.com/problem/reverse-words-in-a-string-iii/description
export default (s) => {
return s.split(' ').map(item => {
let tmp = ''
for (let n = item.length; n >= 0; n--) {
tmp += item.charAt(n)
}
return tmp
}).join(' ')
}

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export const isUnique = (str) => {
return new Set(str.split('')).size === str.length
}

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/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[]}
*/
export const rightSideView = function (root) {
if (!root) return []
const queue = [root] // 队列 把树顶加入队列
const arr = [] // 用来存储每层最后个元素值
while (queue.length > 0) {
let len = queue.length
while (len) {
const node = queue.shift() // 取出队列第一个元素
if (len === 1) arr.push(node.val) // 当是 当前一层的最后一个元素时把值加入arr
if (node.left) queue.push(node.left) // 继续往队列添加元素
if (node.right) queue.push(node.right)
len--
}
}
return arr
}

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import { updateMatrix } from '../../src/array/01-matrix'
test('01 矩阵', () => {
expect(updateMatrix([
[0, 0, 0],
[0, 1, 0],
[0, 0, 0]
])).toEqual([
[0, 0, 0],
[0, 1, 0],
[0, 0, 0]
])
expect(updateMatrix([
[0, 0, 0],
[0, 1, 0],
[1, 1, 1]
])).toEqual([
[0, 0, 0],
[0, 1, 0],
[1, 2, 1]
])
expect(updateMatrix([
[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0]
])).toEqual([
[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0],
[0, 0, 0]
])
expect(updateMatrix([
[0, 1, 0, 1, 1],
[1, 1, 0, 0, 1],
[0, 0, 0, 1, 0],
[1, 0, 1, 1, 1],
[1, 0, 0, 0, 1]
])).toEqual([
[0, 1, 0, 1, 2],
[1, 1, 0, 0, 1],
[0, 0, 0, 1, 0],
[1, 0, 1, 1, 1],
[1, 0, 0, 0, 1]
])
})

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import maxProfit from '../../src/array/best-time-to-buy-and-sell-stock-ii'
test('maxProfit', () => {
expect(maxProfit([7, 1, 5, 3, 6, 4])).toEqual(7)
expect(maxProfit([1, 2, 3, 4, 5])).toEqual(4)
expect(maxProfit([7, 6, 4, 3, 1])).toEqual(0)
expect(maxProfit([])).toEqual(0)
expect(maxProfit([2, 1, 2, 0, 1])).toEqual(2)
})

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import { search } from '../../src/array/binary-search'
test('BinarySearch', () => {
expect(search([-1, 0, 3, 5, 9, 12], 9)).toEqual(4)
expect(search([-1, 0, 3, 5, 9, 12], 2)).toEqual(-1)
expect(search([1, 4, 4, 5, 7, 7, 8, 9, 9, 10], 1)).toEqual(0)
expect(search([1, 2, 3, 3, 4, 5, 10], 3)).toEqual(2)
expect(search([1, 2, 3, 3, 4, 5, 10], 6)).toEqual(-1)
})

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import bubbleSort from '../../src/array/bubble-sort'
test('bubbleSort', () => {
expect(bubbleSort([1, 2, 3])).toEqual([1, 2, 3])
expect(bubbleSort([3, 2, 1])).toEqual([1, 2, 3])
expect(bubbleSort([])).toEqual([])
expect(bubbleSort([3, 6, 4, 5, 6, 8])).toEqual([3, 4, 5, 6, 6, 8])
expect(bubbleSort([1, 1, 1])).toEqual([1, 1, 1])
expect(bubbleSort([-1, -10, 3])).toEqual([-10, -1, 3])
expect(bubbleSort([7, 2, 8, 3, 6, 3])).toEqual([2, 3, 3, 6, 7, 8])
})

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import canPlaceFlowers from '../../src/array/can-place-flowers'
test('canPlaceFlowers', () => {
expect(canPlaceFlowers([1, 0, 0, 0, 1], 1)).toEqual(true)
expect(canPlaceFlowers([1, 0, 0, 0, 1], 2)).toEqual(false)
expect(canPlaceFlowers([1, 0, 0, 0, 1, 0, 0], 2)).toEqual(true)
})

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import circularArrayLoop from '../../src/array/circular-array-loop'
test('circularArrayLoop', () => {
expect(circularArrayLoop([2, -1, 1, 2, 2])).toEqual(true)
expect(circularArrayLoop([-1, 2])).toEqual(false)
expect(circularArrayLoop([-2, 1, -1, -2, -2])).toEqual(false)
})

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import { maxArea } from '../../src/array/container-with-most-water'
test('盛最多水的容器', () => {
expect(maxArea([1, 8, 6, 2, 5, 4, 8, 3, 7])).toBe(49)
expect(maxArea([1, 3, 2])).toBe(2)
expect(maxArea([1, 3, 2, 2])).toBe(4)
})

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import { numberOfSubarrays } from '../../src/array/count-number-of-nice-subarrays'
test('统计「优美子数组」', () => {
expect(numberOfSubarrays([1, 1, 2, 1, 1], 3)).toBe(2)
expect(numberOfSubarrays([2, 4, 6], 1)).toBe(0)
expect(numberOfSubarrays([2, 2, 2, 1, 2, 2, 1, 2, 2, 2], 2)).toBe(16)
})

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import { includesInStr, getStrCopyByNum, getMaxRepetitions } from '../../src/array/count-the-repetitions'
test('判断从 s2 中删除某些字符是否可以变为 s1', () => {
expect(includesInStr('abc', 'ab')).toBe(1)
expect(includesInStr('acb', 'ab')).toBe(1)
expect(includesInStr('ab', 'ab')).toBe(1)
expect(includesInStr('ac', 'ab')).toBe(0)
expect(includesInStr('aa', 'a')).toBe(2)
expect(includesInStr('aaa', 'a')).toBe(3)
expect(includesInStr('abcabc', 'abb')).toBe(1)
})
test('获取重复字符串', () => {
expect(getStrCopyByNum('abc', 2)).toBe('abcabc')
expect(getStrCopyByNum('abc', 3)).toBe('abcabcabc')
})
test('统计重复个数', () => {
expect(getMaxRepetitions('abc', 4, 'ab', 2)).toBe(2)
expect(getMaxRepetitions('acb', 4, 'ab', 2)).toBe(2)
expect(getMaxRepetitions('abc', 4, 'abb', 2)).toBe(1)
expect(getMaxRepetitions('aaa', 3, 'aa', 1)).toBe(4)
expect(getMaxRepetitions('abccab', 4, 'abc', 2)).toBe(2)
})

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import { commonChars } from '../../src/array/find-common-characters'
describe('查找常用字符', () => {
test('查找常用字符', () => {
expect(commonChars(['bella', 'label', 'roller']).sort()).toEqual(['e', 'l', 'l'].sort())
expect(commonChars(['cool', 'lock', 'cook']).sort()).toEqual(['c', 'o'].sort())
expect(commonChars(['acabcddd', 'bcbdbcbd', 'baddbadb', 'cbdddcac', 'aacbcccd', 'ccccddda', 'cababaab', 'addcaccd']).sort()).toEqual([].sort())
})
})

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import firstMissingPositive from '../../src/array/first-missing-positive'
test('缺失的第一个正数', () => {
expect(firstMissingPositive([-1])).toBe(1)
expect(firstMissingPositive([0])).toBe(1)
expect(firstMissingPositive([])).toBe(1)
expect(firstMissingPositive([3, 2, -1])).toBe(1)
expect(firstMissingPositive([3, 4, -1, 1])).toBe(2)
expect(firstMissingPositive([1, 2, 0])).toBe(3)
expect(firstMissingPositive([7, 8, 9, 11, 12])).toBe(1)
})

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import firstMissingPrime, { isPrinme } from '../../src/array/first-missing-prime-number'
test('是否是素数', () => {
expect(isPrinme(0)).toBe(false)
expect(isPrinme(1)).toBe(false)
expect(isPrinme(3)).toBe(true)
expect(isPrinme(5)).toBe(true)
expect(isPrinme(7)).toBe(true)
expect(isPrinme(6)).toBe(false)
})
test('缺失的第一个素数', () => {
expect(firstMissingPrime([3, 5, 7])).toBe(2)
expect(firstMissingPrime([2, 3, 5, 7, 11, 13, 17, 23, 29])).toBe(19)
})

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import grayCode from '../../src/array/gray-code'
test('grayCode', () => {
expect(grayCode(1)).toEqual([0, 1])
expect(grayCode(2)).toEqual([0, 1, 3, 2])
expect(grayCode(0)).toEqual([0])
})

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import { canJump } from '../../src/array/jump-game'
test('跳跃游戏', () => {
expect(canJump([2, 3, 1, 1, 4])).toBe(true)
expect(canJump([3, 2, 1, 0, 4])).toBe(false)
})

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import findKthLargest from '../../src/array/kth-largest-element-in-an-array'
test('findKthLargest', () => {
expect(findKthLargest([3, 2, 1, 5, 6, 4], 2)).toEqual(5)
expect(findKthLargest([3, 2, 3, 1, 2, 4, 5, 5, 6], 4)).toEqual(4)
})

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import letterCombinations from '../../src/array/letter-combinations-of-a-phone-number'
test('letterCombinations', () => {
expect(letterCombinations('23')).toEqual(['ad', 'ae', 'af', 'bd', 'be', 'bf', 'cd', 'ce', 'cf'])
})

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import loopAscArray from '../../src/array/loop-asc-array'
test('loopAscArray', () => {
expect(loopAscArray([0, 3, 4, 6, 7], 0)).toEqual(true)
expect(loopAscArray([6, 7, 0, 3, 4], 0)).toEqual(true)
expect(loopAscArray([6, 7, 0, 3, 4], 5)).toEqual(false)
expect(loopAscArray([1, 2, 3, 4, 5], 6)).toEqual(false)
expect(loopAscArray([1, 2, 3, 4, 5], 1)).toEqual(true)
expect(loopAscArray([3, 4, 5, 1, 1, 2], 1)).toEqual(true)
expect(loopAscArray([3, 4, 5, 1, 1, 2], 6)).toEqual(false)
expect(loopAscArray([3, 4, 5, 1, 1, 2], 0)).toEqual(false)
expect(loopAscArray([1, 0, 1, 1, 1], 0)).toEqual(true)
expect(loopAscArray([1, 0, 1, 1, 1], 2)).toEqual(false)
expect(loopAscArray([3], 2)).toEqual(false)
expect(loopAscArray([2], 2)).toEqual(true)
expect(loopAscArray([1, 1, 1], 2)).toEqual(false)
expect(loopAscArray([1, 1, 1], 1)).toEqual(true)
expect(loopAscArray([1, 1, 1, 0], 1)).toEqual(true)
expect(loopAscArray([1, 1, 1, 0], 2)).toEqual(false)
})

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import { majorityElement } from '../../src/array/majority-element'
test('主元素', () => {
expect(majorityElement([1, 2, 5, 9, 5, 9, 5, 5, 5])).toBe(5)
expect(majorityElement([1, 1, 1, 1, 2, 2, 2])).toBe(1)
expect(majorityElement([1, 1, 1, 2, 2, 2, 2])).toBe(2)
expect(majorityElement([3, 2])).toBe(-1)
expect(majorityElement([2, 2, 1, 1, 1, 2, 2])).toBe(2)
expect(majorityElement([1])).toBe(1)
expect(majorityElement([2, 2])).toBe(2)
})

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import maximumGap from '../../src/array/maximum-gap'
import inData from './maximum-gap.test.data'
test('最大间距', () => {
expect(maximumGap([3, 6, 9, 1])).toBe(3)
expect(maximumGap([10])).toBe(0)
expect(maximumGap([13, 16, 19, 1])).toBe(12)
expect(maximumGap([1, 3, 100])).toBe(97)
expect(maximumGap(inData)).toBe(2147428092)
})

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import { numIslands } from '../../src/array/number-of-islands'
test('岛屿的个数', () => {
expect(numIslands([
[1, 1, 0, 0, 0],
[0, 1, 0, 0, 1],
[0, 0, 0, 1, 1],
[0, 0, 0, 0, 0],
[0, 0, 0, 0, 1]
])).toBe(3)
expect(numIslands([
[1, 1]
])).toBe(1)
expect(numIslands([])).toBe(0)
})

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import partitionArray from '../../src/array/partition-array'
test('loopAscArray', () => {
expect(partitionArray([], 9)).toEqual(0)
expect(partitionArray([3, 2, 2, 1], 2)).toEqual(1)
})

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import removeDuplicates from '../../src/array/remove-duplicates-from-sorted-array'
test('removeDuplicates', () => {
const nums = [1, 1, 2]
expect(removeDuplicates(nums)).toBe(2)
expect(nums).toEqual([1, 2])
})
test('removeDuplicates2', () => {
const nums = [1, 2]
expect(removeDuplicates(nums)).toBe(2)
expect(nums).toEqual([1, 2])
})

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import {
reversePairs,
merge,
mergeSort,
resetCount
} from '../../src/array/reverse-pairs'
describe('逆序对', () => {
beforeEach(() => {
resetCount()
})
test('归并排序 - 合并左右', () => {
expect(merge([7], [5])).toEqual([5, 7])
expect(merge([2], [3])).toEqual([2, 3])
expect(merge([2, 3], [4, 5])).toEqual([2, 3, 4, 5])
})
test('归并排序', () => {
expect(mergeSort([2, 4, 3, 5])).toEqual([2, 3, 4, 5])
expect(mergeSort([2, 4, 3, 0])).toEqual([0, 2, 3, 4])
})
test('逆序对', () => {
expect(reversePairs([7, 5, 6, 4])).toBe(5)
expect(reversePairs([2, 4, 1, 3, 5])).toBe(3)
expect(reversePairs([1, 2, 3, 4])).toBe(0)
})
})

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import { search } from '../../src/array/search-in-rotated-sorted-array'
test('搜索旋转排序数组', () => {
expect(search([4, 5, 6, 7, 0, 1, 2], 0)).toBe(4)
expect(search([4, 5, 6, 7, 0, 1, 2], 3)).toBe(-1)
})

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import searchInsert from '../../src/array/search-insert-position'
test('searchInsert', () => {
expect(searchInsert([1, 3, 5, 6], 5)).toEqual(2)
expect(searchInsert([1, 3, 5, 6], 2)).toEqual(1)
expect(searchInsert([1, 3, 5, 6], 7)).toEqual(4)
expect(searchInsert([1, 3, 5, 6], 0)).toEqual(0)
})

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import selectSort from '../../src/array/select-sort'
test('selectSort', () => {
expect(selectSort([1, 2, 3])).toEqual([1, 2, 3])
expect(selectSort([3, 2, 1])).toEqual([1, 2, 3])
expect(selectSort([])).toEqual([])
expect(selectSort([3, 6, 4, 5, 6, 8])).toEqual([3, 4, 5, 6, 6, 8])
expect(selectSort([1, 1, 1])).toEqual([1, 1, 1])
expect(selectSort([-1, -10, 3])).toEqual([-10, -1, 3])
expect(selectSort([7, 2, 8, 3, 6, 3])).toEqual([2, 3, 3, 6, 7, 8])
})

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import { singleNumbers } from '../../src/array/shu-zu-zhong-shu-zi-chu-xian-de-ci-shu-lcof'
test('面试题56 - I. 数组中数字出现的次数', () => {
expect(singleNumbers([4, 1, 4, 6]).some(i => [1, 6].includes(i))).toBe(true)
expect(singleNumbers([1, 2, 10, 4, 1, 4, 3, 3]).some(i => [2, 10].includes(i))).toBe(true)
})

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import { singleNumber } from '../../src/array/single-number'
test('只出现一次的数字', () => {
expect(singleNumber([2, 2, 1])).toBe(1)
expect(singleNumber([4, 1, 2, 1, 2])).toBe(4)
})

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import sortArrayByParity from '../../src/array/sort-array-by-parity'
test('sortArrayByParity', () => {
expect(sortArrayByParity([4, 2, 5, 7])).toEqual([2, 5, 4, 7])
})

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import hasGroupsSizeX from '../../src/array/x-of-a-kind-in-a-deck-of-cards'
test('hasGroupsSizeX', () => {
expect(hasGroupsSizeX([1, 2, 3, 4, 4, 3, 2, 1])).toEqual(true)
expect(hasGroupsSizeX([1, 1, 1, 2, 2, 2, 3, 3])).toEqual(false)
expect(hasGroupsSizeX([1, 1, 2, 2, 2, 2])).toEqual(true)
expect(hasGroupsSizeX([1])).toEqual(false)
expect(hasGroupsSizeX([1, 1])).toEqual(true)
expect(hasGroupsSizeX([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 10, 10, 10])).toEqual(true)
})

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import { getPrimes, nthSuperUglyNumber } from '../../src/heap/super-ugly-number'
test('计算质因数', () => {
expect(getPrimes(6)).toEqual([2, 3])
expect(getPrimes(4)).toEqual([2])
expect(getPrimes(180)).toEqual([2, 3, 5])
})
// test('superUglyNumber', () => {
// expect(nthSuperUglyNumber(12, [2, 7, 13, 19])).toBe(32)
// expect(nthSuperUglyNumber(800, [37, 43, 59, 61, 67, 71, 79, 83, 89, 97, 101, 103, 113, 127, 131, 157, 163, 167, 173, 179, 191, 193, 197, 199, 211, 229, 233, 239, 251, 257])).toBe(32)
// })

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import { merge, mergeKLists } from '../../src/list/merge-k-sorted-lists'
function ListNode (val) {
this.val = val
this.next = null
}
const arr2List = (arr) => {
let first = null
let res = null
for (let n = 0, len = arr.length; n < len; n++) {
const tmp = new ListNode(arr[n])
if (res) res.next = tmp
else first = tmp
res = tmp
}
return first
}
test('数组转链表', () => {
expect(arr2List([1, 2, 3])).toEqual({ next: { next: { next: null, val: 3 }, val: 2 }, val: 1 })
})
test('合并俩链表', () => {
const source1 = arr2List([1, 4, 5])
const source2 = arr2List([1, 3, 4])
const out = [1, 1, 3, 4, 4, 5]
expect(merge(source1, source2)).toEqual(arr2List(out))
})
test('合并K个排序链表 - 1', () => {
const source = [
arr2List([1, 4, 5]),
arr2List([1, 3, 4]),
arr2List([2, 6])
]
const out = [1, 1, 2, 3, 4, 4, 5, 6]
expect(mergeKLists(source)).toEqual(arr2List(out))
})
test('合并K个排序链表 - 2', () => {
const source = [
arr2List([2, 4, null]),
arr2List([null]),
arr2List([-1, null])
]
const out = [-1, 2, 4, null]
expect(mergeKLists(source)).toEqual(arr2List(out))
})
test('合并K个排序链表 - 3', () => {
const source = [
arr2List([2, 6, null]),
arr2List([5, null]),
arr2List([7, null])
]
const out = [2, 5, 6, 7, null]
expect(mergeKLists(source)).toEqual(arr2List(out))
})

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import { waysToChange } from '../../src/math/coin-lcci'
test('硬币', () => {
expect(waysToChange(5)).toBe(2)
expect(waysToChange(10)).toBe(4)
})

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import trailingZeroes from '../../src/math/factorial-trailing-zeroes'
test('trailingZeroes', () => {
expect(trailingZeroes(3)).toBe(0)
expect(trailingZeroes(5)).toBe(1)
})

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import { permute } from '../../src/math/permutations'
test('全排列', () => {
expect(permute([1, 2, 3])).toEqual([
[1, 2, 3],
[1, 3, 2],
[2, 1, 3],
[2, 3, 1],
[3, 1, 2],
[3, 2, 1]
])
})

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import { nthUglyNumber } from '../../src/math/ugly-number-ii.js'
test('丑数 II', () => {
expect(nthUglyNumber(9)).toBe(10)
expect(nthUglyNumber(1)).toBe(1)
expect(nthUglyNumber(10)).toBe(12)
})

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import { maximalRectangle } from '../../src/stack/maximal-rectangle'
test('maximalRectangle', () => {
expect(maximalRectangle([
['1', '0', '1', '0', '0'],
['1', '0', '1', '1', '1'],
['1', '1', '1', '1', '1'],
['1', '0', '0', '1', '0']
])).toBe(6)
expect(maximalRectangle([
[1, 1, 0, 0, 1],
[0, 1, 0, 0, 1],
[0, 0, 1, 1, 1],
[0, 0, 1, 1, 1],
[0, 0, 0, 0, 1]
])).toBe(6)
expect(maximalRectangle([
[true, true, false, false, true],
[false, true, false, false, true],
[false, false, true, true, true],
[false, false, true, true, true],
[false, false, false, false, true]
])).toBe(6)
expect(maximalRectangle([
[0, 0],
[0, 0]
])).toBe(0)
expect(maximalRectangle([['1']])).toBe(1)
expect(maximalRectangle([['0', '1']])).toBe(1)
expect(maximalRectangle([['1', '1']])).toBe(2)
expect(maximalRectangle([['1', '1', '1', '1']])).toBe(4)
expect(maximalRectangle([[1], [0], [1], [1], [1], [1], [0]])).toBe(4)
expect(maximalRectangle([[1], [0], [1], [1], [1], [1], [1], [0]])).toBe(5)
expect(maximalRectangle([['0', '1'], ['1', '0']])).toBe(1)
expect(maximalRectangle([['0', '1'], ['0', '1']])).toBe(2)
expect(maximalRectangle([['1', '0'], ['1', '0']])).toBe(2)
expect(maximalRectangle([['0', '0', '0'], ['0', '0', '0'], ['1', '1', '1']])).toBe(3)
})

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import { compareStrings } from '../../src/string/compare-strings'
test('比较字符串', () => {
expect(compareStrings('ABCD', 'AB')).toBe(true)
expect(compareStrings('ABCD', 'ACD')).toBe(true)
expect(compareStrings('ABCD', 'AABC')).toBe(false)
})

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import { containsDuplicate } from '../../src/string/contains-duplicate'
test('存在重复元素', () => {
expect(containsDuplicate([1, 2, 3, 1])).toBe(true)
expect(containsDuplicate([1, 2, 3, 4])).toBe(false)
expect(containsDuplicate([1, 1, 1, 3, 3, 4, 3, 2, 4, 2])).toBe(true)
})

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import regularExpressionMatching from '../../src/string/regular-expression-matching'
test('regularExpressionMatching', () => {
expect(regularExpressionMatching('aa', 'a')).toEqual(false)
expect(regularExpressionMatching('aaa', 'aa')).toEqual(false)
expect(regularExpressionMatching('aa', 'aa')).toEqual(true)
expect(regularExpressionMatching('aa', 'a*')).toEqual(true)
expect(regularExpressionMatching('ab', '.*')).toEqual(true)
expect(regularExpressionMatching('aab', 'c*a*b')).toEqual(true)
expect(regularExpressionMatching('mississippi', 'mis*is*p*.')).toEqual(false)
expect(regularExpressionMatching('mississippi', 'mis*is*ip*.')).toEqual(true)
})

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