问题形容：

启动两个线程，

• 线程A打印数组{"A","B","C","D","E","F","G"};
• 线程B打印数组{"1","2","3","4","5","6","7"};

要求，打印进去的后果是

A 1 B 2 C 3 D 4 E 5 F 6 G 7

剖析
用到的知识点：synchronized、notify、wait

代码

public class Demo {    private static String array1[] = {"A","B","C","D","E","F","G"};    private static String array2[] = {"1","2","3","4","5","6","7"};    public static void main(String[] args) {        //定义同一把锁        Object o = new Object();        //线程1        new Thread(() -> {            for (int i = 0; i < array1.length; i++) {                synchronized (o) {                    System.out.print(array1[i] + " ");                    o.notify();                    try {                        o.wait();                    } catch (InterruptedException e) {                        e.printStackTrace();                    }                }            }        }).start();        //线程2        new Thread(() -> {            for (int i = 0; i < array2.length; i++) {                synchronized (o) {                    System.out.print(array2[i] + " ");                    o.notify();                    try {                        o.wait();                    } catch (InterruptedException e) {                        e.printStackTrace();                    }                }            }        }).start();    }}

知识点总结

wait notify notifyAll

// 都是被 `final` 润饰的办法，不能被重写。public final native void notify();public final void wait() throws InterruptedException {    wait(0);}public final native void notifyAll();

• 当线程执行 wait() 办法时，会开释以后的锁，而后让出CPU，进入期待状态。
• 只有当 notify/notifyAll 被执行的时候，才会唤醒一个或多个正在处于期待状态的线程，而后持续往下执行，直到执行完synchronized代码块的代码或是中途遇到wait()，再次开释锁。
• notify() 办法的执行只是唤醒沉睡的线程，而不会立刻开释锁。

notify() 和 notifyAll() 的区别

• notify 办法只唤醒一个期待线程并使该线程开始执行。
• notifyAll 办法会唤醒所有期待线程

经典案例
生产者-消费者问题

Java多线程学习之wait、notify/notifyAll 详解