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题目要求：创建三个线程，每个线程分别打印ABC，并按照ABC的顺序执行十次
题目可以使用多种不同的方式解决，下面我们分别使用 Condition 等待唤醒机制、Semaphore 信号量、CountDownLatch
闭锁、Thread.join() 方法四种方式实现题目要求。
一、使用一个 ReentrantLock 和 三个 Condition 来实现：
import java.util.concurrent.locks.Condition; import
java.util.concurrent.locks.ReentrantLock; /** * 题目要求：ABC三个线程顺序执行10次 *
实现思路：使用一个ReentrantLock 和 三个 Condition 来实现 */ public class
PrintABCUsingCondition { private static ReentrantLock lock = new
ReentrantLock(); private static Condition conditionA = lock.newCondition();
private static Condition conditionB = lock.newCondition(); private static
Condition conditionC = lock.newCondition(); public void execute(String flag) {
lock.lock(); for (int i = 1 ; i <= 10 ; i++){ if ("A".equals(flag)) print(flag,
conditionA, conditionB); if ("B".equals(flag)) print(flag, conditionB,
conditionC); if ("C".equals(flag)) print(flag, conditionC, conditionA); }
lock.unlock(); } private void print(String name, Condition currentThread,
Condition nextThread) { try{
System.out.println(Thread.currentThread().getName() + "-" + name);
nextThread.signal(); currentThread.await(); }catch (InterruptedException e){
e.printStackTrace(); } } public static void main(String[] args) throws
InterruptedException { PrintABCUsingCondition myTask = new
PrintABCUsingCondition(); new Thread(() -> myTask.execute("A")).start();
//必须确保线程A比另外两个线程先拿到ReentrantLock，所以让主线程sleep一段时间 Thread.sleep(500); new
Thread(() -> myTask.execute("B")).start(); new Thread(() ->
myTask.execute("C")).start(); } }
二、基于 Semaphore 信号量来实现：
import java.util.concurrent.Semaphore; /** * 题目要求：ABC三个线程顺序执行10次 *
实现思路：使用一个Semaphore信号量来实现 */ class PrintABCUsingSemaphore { private Semaphore
semaphoreA = new Semaphore(1); private Semaphore semaphoreB = new Semaphore(0);
private Semaphore semaphoreC = new Semaphore(0); private void printA(){
print("A", semaphoreA, semaphoreB); } private void printB(){ print("B",
semaphoreB, semaphoreC); } private void printC(){ print("C", semaphoreC,
semaphoreA); } private void print(String name, Semaphore currentSemaphore,
Semaphore nextSemaphore) { for (int i = 0; i < 10; i++){ try {
currentSemaphore.acquire(); System.out.println(Thread.currentThread().getName()
+" print "+ name); nextSemaphore.release(); } catch (InterruptedException e) {
e.printStackTrace(); } } } public static void main(String[] args) {
PrintABCUsingSemaphore printABC = new PrintABCUsingSemaphore(); new Thread(()
-> printABC.printA()).start(); new Thread(() -> printABC.printB()).start(); new
Thread(() -> printABC.printC()).start(); } }
三、基于 CountDownLatch 闭锁来实现：
import java.util.HashMap; import java.util.Map; import java.util.concurrent.*;
/** * 题目要求：ABC三个线程顺序执行10次 * 实现思路：使用 CountDownLatch 来实现： *
（1）定义dependLatch(所依赖的latch名)、selfLatch(自己的latch名) *
（2）首先调用所依赖的latch的await()方法，如果所依赖的latch的count为0，则重置所依赖的latch并打印需要输出的，最后将自身的count减去
* （3）sum为需要执行的次数 */ public class PrintABCUsingCountDownLatch implements
Runnable { private static Map<String, CountDownLatch> countDownLatchMap = new
HashMap<>(); private String dependLatch; private String selfLatch; private
PrintABCUsingCountDownLatch(String dependLatch, String selfLatch) {
this.dependLatch = dependLatch; this.selfLatch = selfLatch; } @Override public
void run() { for (int i = 0; i < 10; i++) { try {
countDownLatchMap.get(dependLatch).await(); countDownLatchMap.put(dependLatch,
new CountDownLatch(1)); System.out.println(Thread.currentThread().getName() +
":" + selfLatch); countDownLatchMap.get(selfLatch).countDown(); }catch
(InterruptedException e){ e.printStackTrace(); } } } public static void
main(String[] args) { String latchA = "A"; String latchB = "B"; String latchC =
"C"; countDownLatchMap.put(latchA, new CountDownLatch(1));
countDownLatchMap.put(latchB, new CountDownLatch(1));
countDownLatchMap.put(latchC, new CountDownLatch(1));
//创建三个线程，但是此时由于三个CountDownLatch都不为0，所以三个线程都处于阻塞状态 Thread threadA = new
Thread(new PrintABCUsingCountDownLatch(latchC, latchA)); Thread threadB = new
Thread(new PrintABCUsingCountDownLatch(latchA, latchB)); Thread threadC = new
Thread(new PrintABCUsingCountDownLatch(latchB, latchC)); threadA.start();
threadB.start(); threadC.start(); //latchC 阻塞了
latchA；调用latchC的countDown()方法，先让latchC为0，使latchA先运行
countDownLatchMap.get(latchC).countDown(); }
四、 使用 Thread.join() 方法来实现：
/** * 题目要求：ABC三个线程顺序执行10次 * 实现思路：使用 Thread.join() 方法来实现 */ public class
PrintABCUsingJoin { public static void main(String[] args) { Thread t0 = new
Thread(new Work((null))); Thread t1 = new Thread(new Work((t0))); Thread t2 =
new Thread(new Work((t1))); t0.start(); t1.start(); t2.start(); } } class Work
implements Runnable { private Thread beforeThread; public Work(Thread
beforeThread) { this.beforeThread = beforeThread; } @Override public void run()
{ //调用前面线程的join方法 if(beforeThread != null) { try{ beforeThread.join(); }catch
(InterruptedException e) { e.printStackTrace(); } } System.out.println("当前线程："
+ Thread.currentThread().getName()); } }