面试题：实现一个容器，提供两个方法，add，size；写两个线程，线程1添加10个元素到容器中，线程2实现监控元素的个数，当个数到5个时，线程2给出提示并结束

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主要思路是生产者消费者。当A线程打印到第5个时，让B线程的信号量加1；B线程开始执行并结束，结束前让A线程的信号量加1，A线程继续add

import java.util.LinkedList;
import java.util.List;
import java.util.concurrent.Semaphore;

public class TwoAddLearn {
    public static void main(String[] args) {
        Container container = new Container();
        Semaphore semaphoreA = new Semaphore(0);
        Semaphore semaphoreB = new Semaphore(0);
        ThreadA threadA = new ThreadA(container, semaphoreA, semaphoreB);
        ThreadB threadB = new ThreadB(container, semaphoreA, semaphoreB);
        threadA.start();
        threadB.start();
    }
}

class Container {
    private List<Integer> list = new LinkedList<>();

    public void add(Integer a) {
        list.add(a);
    }

    public int getSize() {
        return list.size();
    }
}

class ThreadA extends Thread {
    private Container container;
    private Semaphore semaphoreA;
    private Semaphore semaphoreB;

    ThreadA(Container container, Semaphore semaphoreA, Semaphore semaphoreB) {
        this.container = container;
        this.semaphoreA = semaphoreA;
        this.semaphoreB = semaphoreB;
    }

    @Override
    public void run() {
        super.run();
        for (int i = 1; i <= 10; i++) {
            if (container.getSize() == 5) {
                semaphoreB.release();
                try {
                    semaphoreA.acquire();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
            System.out.println(i);
            container.add(i);
        }
    }
}

class ThreadB extends Thread {
    private Container container;
    private Semaphore semaphoreA;
    private Semaphore semaphoreB;

    ThreadB(Container container, Semaphore semaphoreA, Semaphore semaphoreB) {
        this.container = container;
        this.semaphoreA = semaphoreA;
        this.semaphoreB = semaphoreB;
    }

    @Override
    public void run() {
        super.run();
        try {
            semaphoreB.acquire();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("线程B结束");
        semaphoreA.release();
    }
}

执行结果

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