停车问题
有停车场。在一个停车场，一次只能有一辆车。如果所有停车场都被占用，那么汽车将等待一段时间，如果仍然没有免费停车场，那么它就会离开。
需要使用线程解决（按遗嘱同步）。
这是我的代码：
停车场
class Parking implements Runnable {
private Thread thread;
private String threadName;
static int parkingLots;
static {
parkingLots = 5;
}
Parking(String threadName) {
this.threadName = threadName;
}
public void run() {
if (parkingLots > 0) {
long restTime = (long) (Math.random() * 2000);
try {
parkingLots--;
System.out.println("Car " + threadName + " stands in the parking lot");
Thread.sleep(restTime);
} catch (InterruptedException e) {
}
parkingLots++;
System.out.println("Car " + threadName + " has left parking, it stood there" + ((double)restTime / (double)1000) + " s");
} else
System.out.println("Car " + threadName + " has left parking");
}
public void start() {
if (thread == null) {
thread = new Thread(this, threadName);
thread.start();
}
}
}
主要
public class Main {
public static void main(String[] args) {
ArrayList<Parking> parking = new ArrayList<Parking>();
for (int i = 0; i < 15; i++) {
parking.add(new Parking(String.valueOf(i + 1)));
}
for (Parking i: parking) {
i.start();
}
}
}
我想看到的（当有2个停车场和4辆车时）：
Car 1 stands in the parking lot
Car 2 stands in the parking lot
Car 3 is waiting
Car 4 is waiting
Car 3 has left parking
Car 2 has left parking, it stood there 1.08 s
Car 4 stands in the parking lot
Car 1 has left parking, it stood there 1.71 s
Car 4 has left parking, it stood there 0.83 s
但我得到的（当有2个停车场和4辆车时）：所有第一辆车（1和2）站在停车场，其他车辆（3和4）刚离开，因为没有免费停车场。即使像15辆汽车那样他们仍然无法进入那里。
那么如何让汽车在离开之前等待一段时间呢？如果有免费停车场那么他们就会去那里，否则他们会离开停车场。
答案 0 :(得分：1)
修改了您的代码，看看是否有效。
public class Parking implements Runnable {
private Thread thread;
private String threadName;
static int parkingLots;
static {
parkingLots = 5;
}
Parking(String threadName) {
this.threadName = threadName;
}
public void run() {
long restTime = (long) (Math.random() * 2000);
if (parkingLots > 0) {
checkparking();
} else {
try {
System.out.println("Car " + threadName + " is waiting");
Thread.sleep(restTime);
System.out.println("Car " + threadName + " is checking for free parkinglot");
checkparking();
} catch (InterruptedException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
public void checkparking() {
if (parkingLots > 0) {
long restTime = (long) (Math.random() * 2000);
try {
parkingLots--;
System.out.println("Car " + threadName + " stands in the parking lot");
Thread.sleep(restTime);
} catch (InterruptedException e) {
}
parkingLots++;
System.out.println(
"Car " + threadName + " has left parking, it stood there" + ((double) restTime / (double) 1000) + " s");
} else {
System.out.println(
"Car " + threadName + " has left since there is no parking space");
}
}
public void start() {
if (thread == null) {
thread = new Thread(this, threadName);
thread.start();
}
}
}
public class Main {
public static void main(String[] args) {
ArrayList<Parking> parking = new ArrayList<Parking>();
for (int i = 0; i < 15; i++) {
parking.add(new Parking(String.valueOf(i + 1)));
}
for (Parking i: parking) {
i.start();
}
}
}
输出：
Car 2 stands in the parking lot
Car 1 stands in the parking lot
Car 7 is waiting
Car 5 stands in the parking lot
Car 3 stands in the parking lot
Car 6 is waiting
Car 4 stands in the parking lot
Car 9 is waiting
Car 8 is waiting
Car 10 is waiting
Car 11 is waiting
Car 12 is waiting
Car 13 is waiting
Car 14 is waiting
Car 15 is waiting
Car 4 has left parking, it stood there0.049 s
Car 14 is checking for free parkinglot
Car 14 stands in the parking lot
Car 5 has left parking, it stood there0.366 s
Car 2 has left parking, it stood there0.461 s
Car 12 is checking for free parkinglot
Car 12 stands in the parking lot
Car 15 is checking for free parkinglot
Car 15 stands in the parking lot
Car 1 has left parking, it stood there0.882 s
Car 9 is checking for free parkinglot
Car 9 stands in the parking lot
Car 10 is checking for free parkinglot
Car 10 has left since there is no parking space
Car 3 has left parking, it stood there1.014 s
Car 13 is checking for free parkinglot
Car 13 stands in the parking lot
Car 15 has left parking, it stood there0.937 s
Car 6 is checking for free parkinglot
Car 6 stands in the parking lot
Car 11 is checking for free parkinglot
Car 11 has left since there is no parking space
Car 13 has left parking, it stood there0.344 s
Car 7 is checking for free parkinglot
Car 7 stands in the parking lot
Car 8 is checking for free parkinglot
Car 8 has left since there is no parking space
Car 7 has left parking, it stood there0.054 s
Car 14 has left parking, it stood there1.731 s
Car 9 has left parking, it stood there1.359 s
Car 12 has left parking, it stood there1.877 s
Car 6 has left parking, it stood there1.787 s
答案 1 :(得分：0)
您没有使用和同步，您的类不应该实现Runnable，并且您使用的语义不正确。
例如，启动后汽车无法停放/离开，多线程是指多个线程同时访问并且不会导致状态不一致的情况，在您的情况下，您基本上只是创建线程并执行某些操作。
你的结构应该是这样的：
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;
public class Parking {
private final Lock monitor = new ReentrantLock();
private final Condition lotAvailable = monitor.newCondition();
private List<Car> parkedCars = new ArrayList<>();
private final int maxCapacity;
private int occupied;
public Parking(int maxCapacity) {
this.maxCapacity = maxCapacity;
}
private boolean tryPark(Car car, int maxWaitingMillis) throws InterruptedException{
try {
monitor.lock();
if(occupied >= maxCapacity){
long nanos = lotAvailable.awaitNanos(maxWaitingMillis);
while (occupied >= maxCapacity) {
if (nanos <= 0L)
return false;
nanos = lotAvailable.awaitNanos(nanos);
}
++occupied;
parkedCars.add(car);
return true;
}
++occupied;
parkedCars.add(car);
return true;
}catch (InterruptedException ie){
System.out.println(ie.getMessage());
throw ie;
}finally {
monitor.unlock();
}
}
private void leave(Car car){
try {
monitor.lock();
if(parkedCars.remove(car)) {
--occupied;
lotAvailable.signal();
}
}catch (Exception e){
System.out.println(e.getMessage());
}finally {
monitor.unlock();
}
}
}