使用javascript解一道关于会议日程安排的面试题

这道面试题是从 HarrisonHao 的一篇博文中看到的：原文链接
我看到之后，感觉此题十分有趣，遂自己用 node.js 以不同的思路实现了一遍，实现中使用了 lodash
原题比较长，而且是英文的，就不粘过来了，完整题目和代码可见github

原题大意

你正在准备一场大型的开发者会议，但是有一点点麻烦……
这场会议为期两天，每天上午从九点开始，上午的会议安排到中午12点之前必须结束；
中午12点到下午1点之间是午餐时间，下午1点开始进行下午的会议，到下午5点前必须结束；
现在你有一个清单，上面写明了所有要安排的议题，和每个议题会占用的时间；
清单如下
Writing Fast Tests Against Enterprise Rails 60min
Overdoing it in Python 45min
Lua for the Masses 30min
Ruby Errors from Mismatched Gem Versions 45min
Common Ruby Errors 45min
Rails for Python Developers lightning
Communicating Over Distance 60min
Accounting-Driven Development 45min
Woah 30min
Sit Down and Write 30min
Pair Programming vs Noise 45min
Rails Magic 60min
Ruby on Rails: Why We Should Move On 60min
Clojure Ate Scala (on my project) 45min
Programming in the Boondocks of Seattle 30min
Ruby vs. Clojure for Back-End Development 30min
Ruby on Rails Legacy App Maintenance 60min
A World Without HackerNews 30min
User Interface CSS in Rails Apps 30min
清单中lightning占用5分钟，其他议题都各自注明了占用时间
现在你要写一个程序把清单上的议题安排进四个时间段内。

用 HarrisonHao 的话说，就是要把19个给定大小的萝卜放进4个给定大小的坑里，坑可以没占满，但萝卜不能剩下。

基本的实现思路

先把清单中的议题按照占用时间排序，方便后续操作，然后开始往时间段中安排，每次拿出一个最耗时的议题，放入最空闲的时段中，不断循环，直到所有议题都被安排进去；

首先，我们先看看有哪些萝卜（读取清单并排序）

本次是在 node.js 中实现的，使用 fs 模块读取文件，按行分割后，将每行内容转为一个 Talk 对象，并按占用时间进行排序：

const fs = require('fs');
const _ = require('lodash');

//会议类
class Talk {
    constructor (line) {
        const words = _.compact(line.split(' '));
        const duration = words.pop();
        if (isNaN(duration[0])) {
            words.push(duration)
            this.duration = 5;
        } else {
            this.duration = duration.slice(0, -4);
        }
        this.name= words.join(' ')
    }
}
//获得输入参数
const args = process.argv.splice(2);
const [inputPath, outputPath] = args;

//读取输入文件内容
let input = fs.readFileSync(inputPath, 'utf8')
//分割输入文件内容
input = _.compact(input.split('\r').join().split('\n'));

//生成会议列表
let talks = input.map(line=>new Talk(line));
//按占用时间从小到大排序
talks = _.orderBy(talks, 'duration');

然后我们来看看有哪几个坑……（生成时间段）

Slot 类实例化时接收一个总可用分钟数，并包含三个方法：
add: 将一个议题添加进该时间段，
used: 返回已使用的分钟数，
balance: 返回空闲的分钟数
每天上午有3个小时会议时间即180分钟，下午240分钟，所以我们可以直接用这两个分钟数来生成时间段：

//时间段类
class Slot {
    constructor (limit) {
        this.limit = limit;
        this.list = [];
    };
    //剩余可用分钟数
    balance () {
        return this.limit - this.used()
    };
    //已安排掉的分钟数
    used () {
        let used = 0;
        this.list.forEach((talk)=>{
            used += (talk.duration-0)
        });
        return used;
    };
    add (talk) {
        this.list.push(talk)
    }
}

//生成时间段列表
let slots = [180, 240, 180, 240].map(limit=>new Slot(limit));

核心部分

现在要把萝卜往坑里扔了，之前提到思路是每次把耗时最长的议题扔进最空闲的时间段中，不断循环，直到所有议题(萝卜)都安排进时间段(坑)里；
获取耗时最长的议题很简单，因为我们之前已经给议题排序了，每次只要.pop()弹出最后一条议题就可以了；
获取最空闲的时间段也是一样的原理，用上 Slot 对象的 balance 方法，按照空闲时间倒序排序，排序结果的第一条就是最空闲的时间段；
具体实现：

//分配会议
function arrange (talks, slots) {
    //拿出一个占用时间最长的会议
    const talk = talks.pop();
    //放到时间最富裕的时间段内
    slots = _.orderBy(slots, slot=>slot.balance(), 'desc');
    slots[0].add(talk);
    if (talks.length == 0) return slots;
    return arrange(talks, slots);
};

//分配每个会议到各个时间段内
const result = arrange(talks, slots);

格式化输出

反正原题似乎是要求按一定格式输出的……不过格式这种东西就不细说了，上个代码得了……

//格式化输出
function print (slots) {
    const formatNum = (num)=>('0'+num).slice(-2);

    const am = {}, pm = {};

    [am[1],am[2],pm[1],pm[2]] = _.orderBy(slots, 'limit');

    [1,2].forEach((i)=>{
        let minutes;
        console.log('Track '+i);
        minutes = 9*60
        am[i].list.forEach((talk)=>{
            const h = formatNum(Math.floor(minutes/60));
            const m = formatNum(minutes%60);
            const outputLine = [
                h,':',m,'AM',' ',
                talk.name,' ',talk.duration,'min'
            ].join('');
            console.log(outputLine);
            minutes += talk.duration-0;
        })
        console.log('12:00PM Lunch');
        minutes = 1*60
        pm[i].list.forEach((talk)=>{
            const h = formatNum(parseInt(minutes/60));
            const m = formatNum(minutes%60);
            const outputLine = [
                h,':',m,'PM',' ',
                talk.name,' ',talk.duration,'min'
            ].join('');
            console.log(outputLine);
            minutes += talk.duration-0;
        })
        console.log('05:00PM Networking Event');
    })
}

print(slots);

最终效果

λ node conference.js input.txt
Track 1
09:00AM Clojure Ate Scala (on my project) 45min
09:45AM Pair Programming vs Noise 45min
10:30AM Overdoing it in Python 45min
11:15AM Lua for the Masses 30min
11:45AM Rails for Python Developers lightning, 5min
12:00PM Lunch
01:00PM Ruby on Rails: Why We Should Move On 60min
02:00PM Rails Magic 60min
03:00PM Ruby Errors from Mismatched Gem Versions 45min
03:45PM User Interface CSS in Rails Apps 30min
04:15PM Woah 30min
05:00PM Networking Event
Track 2
09:00AM Writing Fast Tests Against Enterprise Rails 60min
10:00AM Accounting-Driven Development 45min
10:45AM Ruby vs. Clojure for Back-End Development 30min
11:15AM Programming in the Boondocks of Seattle 30min
12:00PM Lunch
01:00PM Ruby on Rails Legacy App Maintenance 60min
02:00PM Communicating Over Distance 60min
03:00PM Common Ruby Errors 45min
03:45PM A World Without HackerNews 30min
04:15PM Sit Down and Write 30min
05:00PM Networking Event

总结

虽然是实现了效果，但我感觉还不完善，现在这种算法还很简单，原题本身就富裕了55分钟的时间，所以才能用这种方式实现，如果占的更满就不一定好使了，以后我会再继续尝试改进；
如果有更好的思路也欢迎告诉我，我也会试着实现一下的~