[CareerCup] 8.3 Musical Jukebox 点唱机
8.3 Design a musical jukebox using object-oriented principles.
CareerCup这书实在是太不负责任了,就写了个半调子的程序,说是完整版也可以下载到,但是我怎么找不到,谁知道在哪里下载请告诉博主,多谢啦~
class Song;
class CD {
public:
// ...
private:
long _id;
string _artist;
set<Song> _songs;
};
class Song {
public:
// ...
private:
long _id;
CD _cd;
string _title;
long _length;
};
class Playlist {
public:
Playlist() {};
Playlist(Song song, queue<Song> queue): _song(song), _queue(queue) {};
Song getNextSToPlay() {
Song next = _queue.front(); _queue.pop();
return next;
}
void queueUpSong(Song s) {
_queue.push(s);
}
private:
Song _song;
queue<Song> _queue;
};
class CDPlayer {
public:
CDPlayer(CD c, Playlist p): _c(c), _p(p) {};
CDPlayer(Playlist p): _p(p) {};
CDPlayer(CD c): _c(c) {};
void playSong(Song s) {}; // ...
Playlist getPlaylist() { return _p; };
void setPlaylist(Playlist p) { _p = p; };
CD getCD() { return _c; };
void setCD(CD c) { _c = c; };
private:
Playlist _p;
CD _c;
};
class User {
public:
User(string name, long id): _name(name), _id(id) {};
string getNmae() { return _name; };
void setName(string name) { _name = name; };
long getID() { return _id; };
void setID(long id) { _id = id; };
User getUser() { return *this; };
static User addUser(string name, long id) {}; // ...
private:
string _name;
long _id;
};
class SongSelector {
public:
Song getCurrentSong() {}; // ...
};
class Jukebox {
public:
Jukebox(CDPlayer cdPlayer, User user, set<CD> cdCollection, SongSelector ts): _cdPlayer(cdPlayer), _user(user), _cdCollection(cdCollection), _ts(ts) {};
Song getCurrentSong() {
return _ts.getCurrentSong();
}
void setUser(User u) {
_user = u;
}
private:
CDPlayer _cdPlayer;
User _user;
set<CD> _cdCollection;
SongSelector _ts;
};
[CareerCup] 8.3 Musical Jukebox 点唱机的更多相关文章
- CareerCup All in One 题目汇总 (未完待续...)
Chapter 1. Arrays and Strings 1.1 Unique Characters of a String 1.2 Reverse String 1.3 Permutation S ...
- CareerCup All in One 题目汇总
Chapter 1. Arrays and Strings 1.1 Unique Characters of a String 1.2 Reverse String 1.3 Permutation S ...
- 《Cracking the Coding Interview》——第8章:面向对象设计——题目3
2014-04-23 18:10 题目:设计一个点唱机. 解法:英文叫Musical Jukebox.这是点唱机么?卡拉OK么?这种题目实在是云里雾里,又没有交流的余地,我索性用一个vector来表示 ...
- POJ1743 Musical Theme [后缀数组]
Musical Theme Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 27539 Accepted: 9290 De ...
- [CareerCup] 18.1 Add Two Numbers 两数相加
18.1 Write a function that adds two numbers. You should not use + or any arithmetic operators. 这道题让我 ...
- [CareerCup] 17.2 Tic Tac Toe 井字棋游戏
17.2 Design an algorithm to figure out if someone has won a game oftic-tac-toe. 这道题让我们判断玩家是否能赢井字棋游戏, ...
- POJ 1743 Musical Theme 二分+后缀数组
Musical Theme Description A musical melody is represented as a sequence of N (1<=N<=20000)no ...
- [CareerCup] 18.12 Largest Sum Submatrix 和最大的子矩阵
18.12 Given an NxN matrix of positive and negative integers, write code to find the submatrix with t ...
- [CareerCup] 18.11 Maximum Subsquare 最大子方形
18.11 Imagine you have a square matrix, where each cell (pixel) is either black or white. Design an ...
随机推荐
- Java基础--常用正则匹配符号(必背,必须背,死都要背)
1.字母:匹配单个字母 (1)A:表示匹配字母A: (2)\\:匹配转义字符“\”: (3)\t:匹配转义字符“\t”: (4)\n:匹配转义字符“\n”: 2.一组字符:任意匹配里面的一个单个字符: ...
- javascript 依次输入自动定焦框
<html> <head> <script type="text/javascript"> function moveNext(object,i ...
- android 小记
1.INSTALL_FAILED_INSUFFICIENT_STORAGE 手机内存不够,删除部分后即可安装.
- node.js之看懂package.json依赖库版本控制
金天:学习一个新东西,就要持有拥抱的心态,如果固守在自己先前的概念体系,就会有举步维艰的感觉.node.js依赖库的版本控制 一般node.js项目会依赖大量第三方module, 那么如何控制modu ...
- web开发相关解决方案
HTML5 API 应用 History.js - gracefully supports the HTML5 History/State APIs pushState + ajax Notify.j ...
- Linux下shell颜色配置
颜色配置涉及以下几个地方(本人常用的):命令提示符,文件及目录名显示,echo -e命令 1.颜色值分为前景色和背景色,颜色码值对应关系如下: Front Back Color 黑 红 绿 黄(棕) ...
- cxf构建webservice的两种方式
一.简介 对于基于soap传输协议的webservice有两种开发模式,代码优先和契约优先的模式.代码优先的模式是通过编写服务器端的代码,使用代码生成wsdl:契约优先模式首先编写wsdl,再通过ws ...
- Sping mvc 环境下使用kaptcha 生成验证码
一.kaptcha 的简介 kaptcha 是一个非常实用的验证码生成工具.有了它,你可以生成各种样式的验证码,因为它是可配置的.kaptcha工作的原理是调用 com.google.code.kap ...
- 2014 Super Training #8 A Gears --并查集
题意: 有N个齿轮,三种操作1.操作L x y:把齿轮x,y链接,若x,y已经属于某个齿轮组中,则这两组也会合并.2.操作Q x y:询问x,y旋转方向是否相同(等价于齿轮x,y的相对距离的奇偶性). ...
- 二分图最大匹配算法-Hopcroft-Karp模板
时间复杂度:O((√V)*E) #include<stdio.h> #include<string.h> ,M=,INF=0x3f3f3f3f; int dx[N],dy[M] ...