题目

1114. 按序打印 - 力扣（Leetcode）

给你一个类：

public class Foo {public void first() { print("first"); }public void second() { print("second"); }public void third() { print("third"); }
}

三个不同的线程 A、B、C 将会共用一个 Foo 实例。

• 线程 A 将会调用 first() 方法
• 线程 B 将会调用 second() 方法
• 线程 C 将会调用 third() 方法

请设计修改程序，以确保 second() 方法在 first() 方法之后被执行，third() 方法在 second() 方法之后被执行。

提示：

• 尽管输入中的数字似乎暗示了顺序，但是我们并不保证线程在操作系统中的调度顺序。
• 你看到的输入格式主要是为了确保测试的全面性。

解法一 

先声明两个AtomicInteger变量并初始化为0，使用他来控制first()和second(),second() 和third()之间的顺序关系。

在进入second和third方法时，提前进入循环去检验变量是否被增加为1，如果是，则进入方法，否则空转。

class Foo {private AtomicInteger firstJobDone = new AtomicInteger(0);private AtomicInteger secondJobDone = new AtomicInteger(0);public Foo() {}public void first(Runnable printFirst) throws InterruptedException {// printFirst.run() outputs "first".printFirst.run();// mark the first job as done, by increasing its count.firstJobDone.incrementAndGet();}public void second(Runnable printSecond) throws InterruptedException {while (firstJobDone.get() != 1) {// waiting for the first job to be done.}// printSecond.run() outputs "second".printSecond.run();// mark the second as done, by increasing its count.secondJobDone.incrementAndGet();}public void third(Runnable printThird) throws InterruptedException {while (secondJobDone.get() != 1) {// waiting for the second job to be done.}// printThird.run() outputs "third".printThird.run();}
}

方法二

CountDownLatch(int count); //构造方法，创建一个值为count 的计数器。
​
//阻塞当前线程，将当前线程加入阻塞队列。
//当计数器的值变为0时，在CountDownLatch上await()的线程就会被唤醒
await();
​
await(long timeout, TimeUnit unit);//在timeout的时间之内阻塞当前线程,时间一过则当前线程可以执行，
​
countDown();//对计数器进行递减1操作，当计数器递减至0时，当前线程会去唤醒阻塞队列里的所有线程。

在构造函数中使用 CountDownLatch（）来说设置两个值为1的计数器，然后在second和third函数中加入awati（）函数进行等待,当计数器为0时被唤醒

代码：

 class Foo {private final CountDownLatch firstDone;private final CountDownLatch secondDone;public Foo() {firstDone = new CountDownLatch(1);secondDone = new CountDownLatch(1);}public void first(Runnable printFirst) throws InterruptedException {// printFirst.run() outputs "first". Do not change or remove this line.printFirst.run();firstDone.countDown();}public void second(Runnable printSecond) throws InterruptedException {firstDone.await();// printSecond.run() outputs "second". Do not change or remove this line.printSecond.run();secondDone.countDown();}public void third(Runnable printThird) throws InterruptedException {secondDone.await();// printThird.run() outputs "third". Do not change or remove this line.printThird.run();}
}

方法三

使用信号量Semaphore ，先初始化信号量，不允许second和third函数运行，在first函数末尾调用release函数使得计数值加一，second函数可以运行；同理，在second函数结束后调用release函数增肌计数值，使得third()函数可以运行。

class Foo {private Semaphore two = new Semaphore(0);private Semaphore three = new Semaphore(0);public Foo() {}public void first(Runnable printFirst) throws InterruptedException {// printFirst.run() outputs "first". Do not change or remove this line.printFirst.run();two.release();}public void second(Runnable printSecond) throws InterruptedException {two.acquire();// printSecond.run() outputs "second". Do not change or remove this line.printSecond.run();three.release();}public void third(Runnable printThird) throws InterruptedException {three.acquire();// printThird.run() outputs "third". Do not change or remove this line.printThird.run();}
}

方法四

使用object和synchronized进行原子操作

public class Foo {private  int flag = 1;private final Object object = new Object();public Foo() {}public void first(Runnable printFirst) throws InterruptedException {synchronized (object) {while (flag != 1) object.wait();printFirst.run();flag = 2;object.notifyAll();}}public void second(Runnable printSecond) throws InterruptedException {synchronized (object) {while (flag != 2) object.wait();printSecond.run();flag = 3;object.notifyAll();}}public void third(Runnable printThird) throws InterruptedException {synchronized (object) {while (flag != 3) object.wait();printThird.run();}}
}