Java-JUC（十二）：有3个线程。线程A和线程B并行执行，线程C需要A和B执行完成后才能执行。可以怎么实现？

方案（一）CountDownLatch：

使用CountDownLatch+Semaphore方式实现：

import java.util.concurrent.CountDownLatch;
import java.util.concurrent.Semaphore;public class TestABC {
    public static void main(String[] args) throws InterruptedException {
        CountDownLatch countDownLatch=new CountDownLatch(2);
        Semaphore semaphoreC = new Semaphore(1);        Thread threadA = new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    Thread.sleep(newjava.util.Random().nextInt(1000));
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName());
                countDownLatch.countDown();
            }
        }, "Thread-A");        Thread threadB = new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    Thread.sleep(newjava.util.Random().nextInt(1000));
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName());
                countDownLatch.countDown();
            }
        }, "Thread-B");        Thread threadC = new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    semaphoreC.acquire();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName());
                semaphoreC.release();
            }
        }, "Thread-C");        // 占用C锁，直到A/B线程完成后，才释放C锁。
        semaphoreC.acquire();        threadA.start();
        threadB.start();
        threadC.start();        countDownLatch.await();
        // 释放C锁，让C线程有获取锁的可能
        semaphoreC.release();    }
}

上边使用CountDownLatch+Semaphore方式实现，但是缺点：上边这种方式会导致线程阻塞情况。下边这种方案是可以实现不阻塞线程的用法：

import java.util.concurrent.CountDownLatch;public class TestABC {
    public static void main(String[] args) throws InterruptedException {
        CountDownLatch countDownLatch=new CountDownLatch(2);        Thread threadA = new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    Thread.sleep(new java.util.Random().nextInt(1000));
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName());
                countDownLatch.countDown();
            }
        }, "Thread-A");        Thread threadB = new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    Thread.sleep(new java.util.Random().nextInt(1000));
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName());
                countDownLatch.countDown();
            }
        }, "Thread-B");        Thread threadC = new Thread(new Runnable() {
            @Override
            public void run() {
                // 在C中等待A/B運算結束
                try {
                    countDownLatch.await();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                    throw new RuntimeException("CountDownLatch等待失败。。。",e);
                }                 System.out.println(Thread.currentThread().getName());
            }
        }, "Thread-C");        threadA.start();
        threadB.start();
        threadC.start();
    }
}

方案（二）：CyclicBarrier

import java.util.concurrent.BrokenBarrierException;
import java.util.concurrent.CyclicBarrier;public class TestABC {
    public static void main(String[] args) throws InterruptedException {
        CyclicBarrier cyclicBarrier=new CyclicBarrier(3);        Thread threadA = new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    Thread.sleep(new java.util.Random().nextInt(1000));
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName());                // 冲破栅栏代表A线程结束
                try {
                    cyclicBarrier.await();
                } catch (InterruptedException | BrokenBarrierException e) {
                    e.printStackTrace();
                    throw new RuntimeException("cylicBarrier.await()拋出異常：",e);
                }
            }
        }, "Thread-A");        Thread threadB = new Thread(new Runnable() {
            @Override
            public void run() {
                try {
                    Thread.sleep(new java.util.Random().nextInt(1000));
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
                System.out.println(Thread.currentThread().getName());                // 冲破栅栏代表B线程结束
                try {
                    cyclicBarrier.await();
                } catch (InterruptedException | BrokenBarrierException e) {
                    e.printStackTrace();
                    throw new RuntimeException("cylicBarrier.await()拋出異常：",e);
                }
            }
        }, "Thread-B");        Thread threadC = new Thread(new Runnable() {
            @Override
            public void run() {
                // 等待前两个(A/B)线程结束，只有前两个(A/B)线程结束了才能满足3个线程都冲破栅栏，
                try {
                    // 等待栅栏被冲破，冲破栅栏的条件是：A/B/C三个线程都到达await()。
                    // 只有栅栏冲破，才能向下执行，否则先到达的线程等待。
                    cyclicBarrier.await();
                } catch (InterruptedException | BrokenBarrierException e) {
                    e.printStackTrace();
                    throw new RuntimeException("cylicBarrier.await()拋出異常：",e);
                }
                // 满足了三个线程都冲破栅栏才向下执行
                System.out.println(Thread.currentThread().getName());
            }
        }, "Thread-C");        threadA.start();
        threadB.start();
        threadC.start();
    }
}