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4 changes: 1 addition & 3 deletions package.json
Original file line number Diff line number Diff line change
Expand Up @@ -6,12 +6,10 @@
"start": "NODE_ENV=production node bin/www",
"dev": "NODE_ENV=development ./node_modules/.bin/nodemon bin/www",
"prd": "pm2 start bin/www",
"test": "echo \"Error: no test specified\" && exit 1",
"md2json": "node ./utils/transforMd2json.js"
"test": "echo \"Error: no test specified\" && exit 1"
},
"dependencies": {
"@koa/cors": "^3.1.0",
"commonmark": "^0.29.3",
"debug": "^4.1.1",
"koa": "^2.7.0",
"koa-bodyparser": "^4.2.1",
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56 changes: 52 additions & 4 deletions routes/dailyProblem.js
Original file line number Diff line number Diff line change
@@ -1,6 +1,5 @@
const router = require("koa-router")();
const solutions = require("../static/solution/solutions.json");
const problems = require("../static/problem/problem.json");
const { decrypt } = require("../utils/crypto");

const { success, fail } = require("../utils/request");
Expand All @@ -18,9 +17,58 @@ router.get("/api/v1/daily-problem", async (ctx) => {
// 3. 根据 Day 几 计算出具体返回哪一个题目
// !!注意: 如果用户指定的时间大于今天,则返回”题目不存在,仅支持查询历史每日一题“

const day = getDay(ctx.query.date || new Date().getTime()); // 用户指定的实际
if (day in problems) {
ctx.body = success(problems[day]);
const date = getDay(ctx.query.date || new Date().getTime()); // 用户指定的实际
if (date === 2) {
ctx.body = success({
day: 2,
title: "821. 字符的最短距离",
link: "https://leetcode-cn.com/problems/plus-one",
tags: ["基础篇", "数组"], // 目前所有 README 都是没有的。因此如果没有的话,你可以先不返回,有的话就返回。后面我慢慢补
pres: ["数组的遍历(正向遍历和反向遍历)"],
description: `
给定一个字符串 S 和一个字符 C。返回一个代表字符串 S 中每个字符到字符串 S 中的字符 C 的最短距离的数组。

示例 1:

输入: S = "loveleetcode", C = 'e'
输出: [3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0]
说明:

- 字符串 S 的长度范围为 [1, 10000]。
- C 是一个单字符,且保证是字符串 S 里的字符。
- S 和 C 中的所有字母均为小写字母。

`,
});
} else if (date <= 1) {
ctx.body = success({
day: 1,
title: "66. 加一",
whys: [
"1. 由于是大家第一次打卡,因此出一个简单题。虽然是简单题,但是如果将加 1 改为加任意的数字,那么就变成了一个非常常见的面试题",
],
link: "https://leetcode-cn.com/problems/plus-one",
tags: ["基础篇", "数组"], // 目前所有 README 都是没有的。因此如果没有的话,你可以先不返回,有的话就返回。后面我慢慢补
pres: ["数组的遍历(正向遍历和反向遍历)"],
description: `
给定一个由整数组成的非空数组所表示的非负整数,在该数的基础上加一。

最高位数字存放在数组的首位, 数组中每个元素只存储单个数字。

你可以假设除了整数 0 之外,这个整数不会以零开头。

示例 1:

输入: [1,2,3]
输出: [1,2,4]
解释: 输入数组表示数字 123。
示例 2:

输入: [4,3,2,1]
输出: [4,3,2,2]
解释: 输入数组表示数字 4321。
`,
});
} else {
ctx.body = fail({
message: "当前暂时没有每日一题,请联系当前讲师进行处理~",
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192 changes: 0 additions & 192 deletions utils/transforMd2json.js

This file was deleted.