Design a parking lot.

see CC150 OO Design for details.
1) n levels, each level has m rows of spots and each row has k spots.So each level has m x k spots.
2) The parking lot can park motorcycles, cars and buses
3) The parking lot has motorcycle spots, compact spots, and large spots
4) Each row, motorcycle spots id is in range[0,k/4)(0 is included, k/4 is not included), compact spots id is in range [k/4,k/4*3) and large spots id is in range [k/4*3,k).
5) A motorcycle can park in any spot
6) A car park in single compact spot or large spot
7) A bus can park in five large spots that are consecutive and within same row. it can not park in small spots

Have you met this question in a real interview?

Yes
Example

level=1, num_rows=1, spots_per_row=11
parkVehicle("Motorcycle_1") // return true
parkVehicle("Car_1") // return true
parkVehicle("Car_2") // return true
parkVehicle("Car_3") // return true
parkVehicle("Car_4") // return true
parkVehicle("Car_5") // return true
parkVehicle("Bus_1") // return false
unParkVehicle("Car_5")
parkVehicle("Bus_1") // return true

CareerCup上的原题,请参见我之前的博客8.4 Parking Lot 停车场问题

LintCode的这道题的C++的OJ应该有问题,因为我的代码在本地调试都正确,不知道为何通过不了OJ,有没有大神持有同样的观点?

// enum type for Vehicle
enum class VehicleSize {
Motorcycle,
Compact,
Large
}; class Vehicle {
public:
virtual VehicleSize size() {}
virtual int spot_num() {}
}; class Bus: public Vehicle {
public:
VehicleSize size() {
return VehicleSize::Large;
}
int spot_num() {
return ;
}
}; class Car: public Vehicle {
public:
VehicleSize size() {
return VehicleSize::Compact;
}
int spot_num() {
return ;
}
}; class Motorcycle: public Vehicle {
public:
VehicleSize size() {
return VehicleSize::Motorcycle;
}
int spot_num() {
return ;
}
}; class Level {
public:
Level(int num_rows, int spots_per_row) {
int moto = spots_per_row / ;
moto_spots.resize(moto);
int car = spots_per_row / * ;
compact_spots.resize(car - moto);
large_spots.resize(spots_per_row - car);
} bool park_vehicle(Vehicle* vehicle, VehicleSize size, int num) {
if (size == VehicleSize::Motorcycle) {
for (int i = ; i < moto_spots.size(); ++i) {
if (moto_spots[i] == NULL) {
moto_spots[i] = vehicle;
vehicle_to_spot[vehicle][VehicleSize::Motorcycle] = i;
return true;
}
}
return false;
} else if (size == VehicleSize::Compact) {
for (int i = ; i < compact_spots.size(); ++i) {
if (compact_spots[i] == NULL) {
compact_spots[i] = vehicle;
vehicle_to_spot[vehicle][VehicleSize::Compact] = i;
return true;
}
}
for (int i = ; i < large_spots.size(); ++i) {
if (large_spots[i] == NULL) {
large_spots[i] = vehicle;
vehicle_to_spot[vehicle][VehicleSize::Large] = i;
return true;
}
}
return false;
} else if (size == VehicleSize::Large) {
for (int i = ; i < large_spots.size(); ++i) {
if (large_spots[i] == NULL) {
bool can_park = true;
for (int j = i; j < i + num; ++j) {
if (large_spots[j] != NULL) {
can_park = false;
break;
}
}
if (can_park) {
for (int j = i; j < i + num; ++j) {
large_spots[j] = vehicle;
}
vehicle_to_spot[vehicle][VehicleSize::Large] = i;
return true;
}
}
}
return false;
}
} void unpark_vehicle(Vehicle *vehicle) {
map<VehicleSize, int> spot = vehicle_to_spot[vehicle];
VehicleSize size = vehicle->size();
if (spot.count(size)) {
int idx = spot[size];
if (size == VehicleSize::Motorcycle) {
moto_spots[idx] = NULL;
} else if (size == VehicleSize::Compact) {
compact_spots[idx] = NULL;
} else {
for (int i = idx; i < large_spots.size(); ++i) {
if (large_spots[i] == vehicle) {
large_spots[i] = NULL;
} else {
break;
}
}
}
} else if (size == VehicleSize::Compact && spot.count(VehicleSize::Large)) {
int idx = spot[VehicleSize::Large];
large_spots[idx] = NULL;
}
} private:
vector<Vehicle*> moto_spots;
vector<Vehicle*> compact_spots;
vector<Vehicle*> large_spots;
map<Vehicle*, map<VehicleSize, int>> vehicle_to_spot;
}; class ParkingLot {
public:
// @param n number of leves
// @param num_rows each level has num_rows rows of spots
// @param spots_per_row each row has spots_per_row spots
ParkingLot(int n, int num_rows, int spots_per_row) {
for (int i = ; i < n; ++i) {
Level *level = new Level(num_rows, spots_per_row);
levels.push_back(level);
}
} // Park the vehicle in a spot (or multiple spots)
// Return false if failed
bool parkVehicle(Vehicle &vehicle) {
for (int i = ; i < levels.size(); ++i) {
if (levels[i]->park_vehicle(&vehicle, vehicle.size(), vehicle.spot_num())) {
vehicle_to_level[&vehicle] = levels[i];
return true;
}
}
return false;
} // unPark the vehicle
void unParkVehicle(Vehicle &vehicle) {
Level *level = vehicle_to_level[&vehicle];
if (level) {
level->unpark_vehicle(&vehicle);
}
}
private:
vector<Level*> levels;
map<Vehicle*, Level*> vehicle_to_level;
};

参考资料:

http://www.jianshu.com/p/2bd60b69393d

[LintCode] Parking Lot 停车场问题的更多相关文章

  1. [CareerCup] 8.4 Parking Lot 停车场问题

    8.4 Design a parking lot using object-oriented principles. LintCode上的原题,请参见我的另一篇博客Parking Lot 停车场问题. ...

  2. English trip V1 - 2.Don't Do That Teacher:Patrick Key: 祈使句(imperatives)

    什么是祈使句?    What's imperatives? 求或者希望别人做什么事或者不做什么事时用的句子:带有命令的语气 In this lesson you will learn how to ...

  3. 新概念英语三 新东方主讲Lesson1

    新概念二 Lesson95 词汇 ①get a shock 吓了一跳,得到一个惊喜 例:his wife got a shock get into a such mess 这么不幸搞得一片狼籍弄得这样 ...

  4. [LintCode]——目录

    Yet Another Source Code for LintCode Current Status : 232AC / 289ALL in Language C++, Up to date (20 ...

  5. [PA2014]Parking

    [PA2014]Parking 题目大意: 停车场是一个宽度为\(w(w\le10^9)\)的矩形.我们以其左下角顶点为原点,坐标轴平行于矩形的边,建立直角坐标系.停车场很长,我们可以认为它一直向右边 ...

  6. bzoj3718 [PA2014]Parking

    Description 你的老板命令你将停车场里的车移动成他想要的样子.停车场是一个长条矩形,宽度为w.我们以其左下角顶点为原点,坐标轴平行于矩形的边,建立直角坐标系.停车场很长,我们可以认为它一直向 ...

  7. C语言实现简单的停车场管理系统

    问题描述:停车场是一个能放n辆车的狭长通道,只有一个大门,汽车按到达的先后次序停放.若车场满了,车要停在门外的便道上等候,一旦有车走,则便道上第一辆车进入.当停车场中的车离开时,由于通道窄,在它后面呢 ...

  8. (lintcode全部题目解答之)九章算法之算法班题目全解(附容易犯的错误)

    --------------------------------------------------------------- 本文使用方法:所有题目,只需要把标题输入lintcode就能找到.主要是 ...

  9. Lintcode 85. 在二叉查找树中插入节点

    -------------------------------------------- AC代码: /** * Definition of TreeNode: * public class Tree ...

随机推荐

  1. 消息队列通信,王明学learn

    消息队列通信 消息队列就是一个消息(一个结构)的链表.而一条消息则可看作一个记录,具有特定的格式.进程可以从中按照一定的规则添加新消息:另一些进程则可以从消息队列中读走消息. 每一个消息都是一个结构体 ...

  2. canvas加载进度条

    <!DOCTYPE html> <html><head><meta http-equiv="Content-Type" content=& ...

  3. 【SSH】 之 Struts

    (一)什么是Struts,Struts是什么? Struts是学习轻量级J2EE框架所必须要了解的一个框架,也是我们当前最最流行的三大框架(SSH——Struts,Spring,Hibernate)之 ...

  4. Hibernate的持久化类状态

    Hibernate的持久化类状态 持久化类:就是一个实体类 与 数据库表建立了映射. Hibernate为了方便管理持久化类,将持久化类分成了三种状态. 瞬时态 transient (临时态):持久化 ...

  5. 【java IO】使用Java输入输出流 读取txt文件内数据,进行拼接后写入到另一个文件中

    package com.sxd.test.util; import java.io.BufferedReader; import java.io.BufferedWriter; import java ...

  6. python 定义类方法

    定义类方法 和属性类似,方法也分实例方法和类方法. 在class中定义的全部是实例方法,实例方法第一个参数 self 是实例本身. 要在class中定义类方法,需要这么写: class Person( ...

  7. Redis的几个认识误区

    此文作者是新浪微博平台架构师杨卫华timyang大师,是他对Redis研究的一个总结,澄清了一些认识上的误区,值得一读. James Hamilton在On Designing and Deployi ...

  8. XSS 跨站脚本攻击之ShellCode的调用

    1.ShellCode,最初是溢出程序和蠕虫病毒的核心,实际上是指利用一个漏洞是所执行的代码,在XSS跨站脚本中,是指由javascript等脚本编写的XSS利用代码: 2.Exploit,在黑客眼里 ...

  9. 【hdu3948-不同回文串的个数】后缀数组

    题意:求不同回文串的个数 n<=10^5 题解: 先按照manacher的构造方法改造一遍串,然后跑一遍manacher. 如ababa--> $#a#b#a#b#a#@ 然后跑一遍后缀数 ...

  10. 解决Genymotion无法创建新设备或无法显示设备列表问题

    准备工作: 链接: https://pan.baidu.com/s/1i5v4IBN 密码: jc3m 用2.8的和最新VirtualBox-5.1.10-112026-Win 注意事项: 1.笔记本 ...