线程交替输出A1B2C3.........Z26

为了完成作业。。。。直接贴代码，复制可直接运行

第一版本

public  static void main(String[] args){
    Object o=new Object();
    Thread t1=new Thread(()->{
        synchronized (o){ //wait 必须在有锁的情况下使用
            for (int i=0;i<26;i++){
                System.out.print((char) (i + 'A'));
                o.notify(); //唤醒其他线程
                try {
                    o.wait(); //自身等待 
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
            o.notify(); //必不可少，不然不会结束
        }
    });

    Thread t2=new Thread(()->{
        synchronized (o){
            for (int i=1;i<=26;i++){
                System.out.print(i);
                o.notify();
                try {
                    o.wait();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
            o.notify();
        }
    });
    t1.start();
    t2.start();
}

　　

经过数次尝试，会出现1A2B3C 的情况；
第二版本

public  static void main(String[] args){
    Object o=new Object();
    CountDownLatch countDownLatch=new CountDownLatch(1); //门闩 1表示等待的数量
    Thread t1=new Thread(()->{
        countDownLatch.countDown();  //唤醒 把countDownLatch的值减1，减到0，就唤醒所有等待在这个countDownLatch上的线程
        synchronized (o){
            for (int i=0;i<26;i++){
                System.out.print((char) (i + 'A'));
                o.notify();
                try {
                    o.wait();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
            o.notify();
        }

    });

    Thread t2=new Thread(()->{
        try {
            countDownLatch.await()；//进入等待状态
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        synchronized (o){
            for (int i=1;i<=26;i++){
                System.out.print(i);
                o.notify();
                try {
                    o.wait();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            }
            o.notify();
        }
    });

    t1.start();
    t2.start();

}

　　

第三种

public static void main(String[] args) {

    t1 = new Thread(() -> {
        for (int i = 0; i < 26; i++) {
            System.out.print((char) (i + 'A'));
            LockSupport.unpark(t2); //唤醒t2 
            LockSupport.park();//暂停当前线程
        }
    });

    t2 = new Thread(() -> {
        for (int i = 1; i <= 26; i++) {
            LockSupport.park();   //这个一定要在打印前暂停
            System.out.print(i); 
            LockSupport.unpark(t1);
        }
    });

    t1.start();
    t2.start();

}

　　

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标签：Z26,synchronized,Thread,int,Object,A1B2C3,线程,countDownLatch,new
来源： https://www.cnblogs.com/count0/p/16698090.html