feat(algorithms): 15 3 sum

Author: guocaoyi

.gitignore

@@ -3,6 +3,7 @@
 .deploy_git
 _book
 node_modules
+.idea
 
 ## files
 **/*.js

CONTRIBUTING.md

@@ -0,0 +1,19 @@
+# CONTRIBUTING
+
+## Running
+
+1. 创建 Problem 目录、index.ts
+2. tsc watch 实时编译
+3. 跑测试用例
+
+# Booking
+
+1. 初始化 README.md
+2. ts 代码转译 AST，获取注释信息
+3. 生成 readme.md 文档
+4. gitbook build
+
+# Deploy
+
+1. git push
+2. git push branch

algorithms/0015.3sum/README.md

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+# 3Sum
+
+# Category Difficulty Likes Dislikes
+
+algorithms Medium (23.92%) 3721 405
+Tags
+Companies
+Given an array nums of n integers, are there elements a, b, c in nums such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
+
+Note:
+
+The solution set must not contain duplicate triplets.
+
+Example:
+
+Given array nums = [-1, 0, 1, 2, -1, -4],
+
+A solution set is:
+[
+[-1, 0, 1],
+[-1, -1, 2]
+]

algorithms/0015.3sum/index.ts

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+/**
+ * 排列组合
+ * 刚拿到这题第一个想法就是使用排列组合，很快写完。结果在遇到一个长3000的数组时，运行超时了，时间复杂度O(n!)。
+ * n!/m!(n-m)! 差不多就是 44.95501亿次运算循环，这不管使用何种语言都会遇到超时的问题
+ * 看了几个 Submission,他们的时间复杂度能做到 O(n^2)
+ * @time 2019.05.20 18:00
+ * @status Time Limit Exceeded
+ * @case [82597,-9243,62390,83030,-97960,-26521...] (3000)
+ */
+var threeSum = (nums: number[]): number[][] => {
+ const result: { [T: string]: number[] } = {};
+ for (let a: number = 0; a < nums.length; a++) {
+ for (let b: number = a + 1; b < nums.length; b++) {
+ for (let c: number = b + 1; c < nums.length; c++) {
+ if (nums[a] + nums[b] + nums[c] === 0) {
+ const arr: number[] = [nums[a], nums[b], nums[c]].sort();
+ const arrStr: string = arr.toString();
+ if (!(arrStr in result)) {
+ result[arrStr] = arr;
+ }
+ }
+ }
+ }
+ }
+ return Object.values(result);
+};
+
+/**
+ * 排列组合
+ * @github https://github.com/guocaoyi/leetcode-ts
+ * @author yalda
+ */
+var threeSum = (nums: number[]): number[][] => {
+ const result: { [T: string]: number[] } = {};
+ nums = nums.sort();
+ let left = 0;
+ let right = nums[nums.length - 1];
+
+ for (let a: number = 0; a < nums.length; a++) {
+ for (let b: number = a + 1; b < nums.length; b++) {
+ for (let c: number = b + 1; c < nums.length; c++) {
+ if (nums[a] + nums[b] + nums[c] === 0) {
+ const arr: number[] = [nums[a], nums[b], nums[c]].sort();
+ const arrStr: string = arr.toString();
+ if (!(arrStr in result)) {
+ result[arrStr] = arr;
+ }
+ }
+ }
+ }
+ }
+
+ return Object.values(result);
+};

package.json

@@ -18,7 +18,7 @@
  "scripts": {
  "clean": "",
  "create": "",
- "watch": "",
+ "watch": "tsc -w ./algorithms/0015.3sum/index.ts",
  "doc": "",
  "index": "",
  "deploy": ""
@@ -44,4 +44,4 @@
  "branch": "gh-pages",
  "dir": "_book"
  }
-}
+}

script/compiler.ts

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+/**
+ * watch & ts compiler
+ */