现在有T1、T2、T3三个线程，怎样保证T2在T1执行完后执行，T3在T2执行完后执行？使用Join

public class TestJoin
{
    public static void main(String[] args)
    {
        Thread t1 = new MyThread("线程1");
        Thread t2 = new MyThread("线程2");
        Thread t3 = new MyThread("线程3");

        try
        {
            //t1先启动
            t1.start();
            t1.join();
            //t2
            t2.start();
            t2.join();
            //t3
            t3.start();
            t3.join();
        }
        catch (InterruptedException e)
        {
            e.printStackTrace();
        }
    }
}

class MyThread extend Thread{
    public MyThread(String name){
        setName(name);
    }
    @Override
    public void run()
    {
        for (int i = 0; i < 5; i++)
        {
            System.out.println(Thread.currentThread().getName()+": "+i);
            try
            {
                Thread.sleep(100);
            }
            catch (InterruptedException e)
            {
                e.printStackTrace();
            }
        }
    }
}

还有一种方式，在t3开始前join t2，在t2开始前join t1

public class TestJoin2
{
    public static void main(String[] args)
    {
        final Thread t1 = new Thread(new Runnable() {

            @Override
            public void run() {
                System.out.println("t1");
            }
        });
        final Thread t2 = new Thread(new Runnable() {

            @Override
            public void run() {
                try {
                    //引用t1线程，等待t1线程执行完
                    t1.join();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("t2");
            }
        });
        Thread t3 = new Thread(new Runnable() {

            @Override
            public void run() {
                try {
                    //引用t2线程，等待t2线程执行完
                    t2.join();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println("t3");
            }
        });
        t3.start();
        t2.start();
        t1.start();
    }
}