经典笔试题：监控容器元素的数量（采用wait和notify实现）

笔试题：
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时，线程2给出提示并结束

import java.util.ArrayList;
import java.util.List;

public class Container {
    List<Object> container = new ArrayList<>();

    public void add(Object o) {
        this.container.add(o);
    }

    public int size() {
        return this.container.size();
    }
}

import java.util.concurrent.TimeUnit;

public class Test {

    public static void main(String[] args) {
        Container container = new Container();

        Object lock = new Object();

        new Thread(() -> {
            System.out.println("t1启动");
            synchronized (lock) {
                for (int i = 0; i < 10; i++) {
                    container.add(new Object());
                    System.out.println("add " + i);

                    if (container.size() == 5) {
                        lock.notify();

                        try {
                            lock.wait();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }

                    }

                }
            }

            System.out.println("t1 结束");
        }, "t1").start();

        try {
            TimeUnit.SECONDS.sleep(1);
        } catch (InterruptedException e1) {
            e1.printStackTrace();
        }

        new Thread(() -> {
            System.out.println("t2启动");
            synchronized (lock) {
                if (container.size() != 5) {
                    try {
                        lock.wait();
                    } catch (InterruptedException e) {
                        e.printStackTrace();
                    }
                }

                System.out.println("size = 5");
                lock.notify();
            }

            System.out.println("t2 结束");
        }, "t2").start();

    }

}

运行结果：

t1启动
add 0
add 1
add 2
add 3
add 4
t2启动
size = 5
t2 结束
add 5
add 6
add 7
add 8
add 9
t1 结束