本篇文章为大家展示了java多线程题目的解决方案，内容简明扼要并且容易理解，绝对能使你眼前一亮，通过这篇文章的详细介绍希望你能有所收获。
在iteye上看到的一道多线程的题目，参考了一下网友的实现，那Eclipse调试通过，算是对JAVA5的并发库有个大致的了解，分享出来，欢迎园里的同学拍砖。
题目：
要求用三个线程，按顺序打印1,2,3,4,5.... 71,72,73,74, 75.
线程1先打印1,2,3,4,5, 然后是线程2打印6,7,8,9,10, 然后是线程3打印11,12,13,14,15. 接着再由线程1打印16,17,18,19,20....以此类推, 直到线程3打印到75。
分析：感觉出题人是要考察一下你是否能够很好的控制多线程，让他们有序的进行。
线程池：3个线程，需要使用并发库的线程池
锁（lcok）：在打印的时候，只允许一个线程进入，其他的线程等待
下面的主要的代码：
import java.util.HashMap; import java.util.Map; import java.util.concurrent.CountDownLatch; import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import java.util.concurrent.locks.Condition; import java.util.concurrent.locks.Lock; import java.util.concurrent.locks.ReentrantLock; public class NumberPrinter { private Lock lock = new ReentrantLock(); private Condition c1 = lock.newCondition(); private Condition c2 = lock.newCondition(); private Condition c3 = lock.newCondition(); private Map<Integer, Condition> condtionContext = new HashMap<Integer, Condition>(); public NumberPrinter() { condtionContext.put(Integer.valueOf(0), c1); condtionContext.put(Integer.valueOf(1), c2); condtionContext.put(Integer.valueOf(2), c3); } private int count = 0; public void print(int id) { lock.lock(); try { while(count*5 < 75) { int curID = calcID(); if (id == curID) { for (int i = 1; i<=5; i++) { System.out.print(count*5 +i+ ","); } System.out.println(); count++; int nextID = calcID(); Condition nextCondition = condtionContext.get( Integer.valueOf(nextID)); //通知下一线程 nextCondition.signal(); } else { Condition condition = condtionContext.get( Integer.valueOf(id)); condition.await(); } } //通知线程结束 for(Condition c : condtionContext.values()) { c.signal(); } } catch (Exception e) { e.printStackTrace(); } finally { lock.unlock(); } } private int calcID() { // TODO Auto-generated method stub return count % 3; } public static void main(String[] args) { ExecutorService executor = Executors.newFixedThreadPool(3); final CountDownLatch latch = new CountDownLatch(1); final NumberPrinter printer = new NumberPrinter(); for (int i = 0; i < 3; i++) { final int id = i; executor.submit(new Runnable() { @Override public void run() { // TODO Auto-generated method stub try { latch.await(); } catch (InterruptedException e) { // TODO Auto-generated catch block e.printStackTrace(); } printer.print(id); } }); } System.out.println("三个任务开始顺序打印数字。。。。。。"); latch.countDown(); executor.shutdown(); } }上述内容就是java多线程题目的解决方案，你们学到知识或技能了吗？如果还想学到更多技能或者丰富自己的知识储备，欢迎关注编程网行业资讯频道。
