练习题10-

1、模拟上车案例:有一辆车,有100个座位,有前中后三个门每个门都可以随机上人，上人的个数不确定。

package com.xxx;

import java.util.Random;

public class CarRunnable implements Runnable {
    private static int seat = 100; //空座位数
    private static int passenger = 0; //乘客数量
    private static String lock = "锁";

    @Override
    public void run() {

        while (true) {
            synchronized (lock) {
                if (seat > 0) {
                    passenger++;
                    seat--;
                    System.out.println("第" + passenger + "位乘客从" + Thread.currentThread().getName() + "上车，座位剩余：" + seat);
                } else {
                    break;
                }
            }
            try {
                //睡一下给其他线程留机会
                //给从其他门进来的乘客抢座的机会
                Thread.sleep(10);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
//            Thread.yield();
        }
    }
}

package com.xxx;

public class Homework1 {
    public static void main(String[] args) {
        CarRunnable door = new CarRunnable();

        Thread t1 = new Thread(door,"①号门");
        Thread t2 = new Thread(door,"②号门");
        Thread t3 = new Thread(door,"③号门");
        t1.start();
        t2.start();
        t3.start();

    }
}

2、实现两个线程打印交替打印1-100

package com.xxx;

public class PrintRunnable implements Runnable {
    private int num = 1;

    @Override
    public void run() {

        while (true) {
            synchronized (this) {

                notify();
                if (num <= 100) {
                    System.out.println(Thread.currentThread().getName() + "：" + num);
                    num++;

                    try {
//                        Thread.sleep(100);
                        this.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                } else {
                    break;
                }
            }
        }
    }
}

package com.xxx;

public class Homework2 {
    public static void main(String[] args) throws InterruptedException {

        PrintRunnable print = new PrintRunnable();
        new Thread(print,"1号打印机").start();
        new Thread(print,"2号打印机").start();

    }
}

3、模拟多线程中的生产者与消费者模式

package com.xxx.homework3;

public class BreadShop {
    private int bread = 10;

    /*
    生产面包
     */
    public synchronized void makeBread() {
        if (bread <= 10) {//假设面包库存10个
            bread++;
            System.out.println(Thread.currentThread().getName() + "制作一个面包》》，店里目前有 " + bread + " 个面包");
            notify();//唤醒吃面包的线程，面包做好了
        } else {
            try {
                wait();//面包库存够了，不用制作面包
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
    }

    /*
    吃面包
     */
    public synchronized void eatBread() {
        if (bread > 0) {
            bread--;
            System.out.println(Thread.currentThread().getName() + "吃掉一个面包**，店里还剩 " + bread + " 个面包");
            notify();//唤醒制作面包的线程
        } else {
            try {
                wait();//等一会儿，面包还没做好
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
        }
    }
}

package com.xxx.homework3;

public class Customer implements Runnable {
    private BreadShop breadShop;

    public Customer(BreadShop breadShop) {
        this.breadShop = breadShop;
    }

    @Override
    public void run() {
        System.out.println(Thread.currentThread().getName() + "要吃面包");
        while (true) {
            try {
                Thread.sleep(500);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            synchronized (this) {
                breadShop.eatBread();

            }
        }
    }
}

package com.xxx.homework3;

public class Producer implements Runnable {
    private BreadShop breadShop;

    public Producer(BreadShop breadShop) {
        this.breadShop = breadShop;
    }

    @Override
    public void run() {
        System.out.println(Thread.currentThread().getName() + "开始做面包》》》");
        while (true) {
            try {
                Thread.sleep(200);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            breadShop.makeBread();
        }
    }
}

package com.xxx.homework3;

public class Test {
    public static void main(String[] args) {
        BreadShop breadShop = new BreadShop();
        Customer customer = new Customer(breadShop);
        Producer producer = new Producer(breadShop);

        new Thread(producer,"生产者").start();
        new Thread(customer,"消费者1").start();
        new Thread(customer,"消费者2").start();
        new Thread(customer,"消费者3").start();

    }
}

4、使用多线程的的两种方式求出1-100偶数之和

package com.xxx.homework4;


public class Summation implements Runnable{

    private int sum = 0;
    private int num = 0;
    /*
    求和线程
     */
    @Override
    public void run() {
        if (num <= 100) {
            synchronized (this) {
                for (int i = 0; i <= 100; i+=2) {
                    if (i % 2 == 0) {
                        sum += i;
                        num++;
                    }
                }
            }
            System.out.println(Thread.currentThread().getName() + "-> 100内偶数和：" + sum);
        } else {
            return;
        }
    }
}
package com.xxx.homework4;

public class Test {
    public static void main(String[] args) {

        Summation summation = new Summation();
        new Thread(summation,"偶数线程").start();

    }
}
package com.xxx.homework4;

public class Test2 {
    private static int oddSum = 0;
    private static int evenSum = 0;

    public static void main(String[] args) {
//        final int sum = 0;
        new Thread(new Runnable() {
            @Override
            public void run() {
                for (int i = 0; i <= 100; i+=2) {
                    oddSum+=i;
                }
                System.out.println(Thread.currentThread().getName() + "->求和" + oddSum);
            }
        },"偶数线程").start();

        new Thread(new Runnable() {
            @Override
            public void run() {
                for (int j = 1; j <= 100; j+=2) {
                    evenSum+=j;
                }
                System.out.println(Thread.currentThread().getName() + "->求和" + evenSum);
            }
        },"奇数线程").start();
    }
}

原文地址：http://www.cnblogs.com/wyzel/p/16819981.html