三个Java线程（使用线程池）交替打印1到75个数的两种解法

解法一：

[java] view
plaincopy

还有另外一个解法：

package net.liuyx.test;



import java.util.ArrayList;

import java.util.List;

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 Test {

private static final int TASK_NUM = 3;

private static int num = 0;

private static int flag = 0;

private static Lock lock = new ReentrantLock();

private static List<Condition> list = new ArrayList<Condition>();

private static ExecutorService exec = Executors.newCachedThreadPool();

static {

for(int i = 0; i < TASK_NUM; i++){

list.add(lock.newCondition());

}

}



private static void crit() {

if (num >= 75) {

System.exit(1);

}

}



private static void print() {

crit();

System.out.print(Thread.currentThread());

for (int i = 0; i < 5; i++) {

System.out.format("%-2d ", ++num);

}

System.out.println();

}



private static void work(int i) {

while (!Thread.interrupted()) {

try{

lock.lock();

if(flag == i){

crit();

print();

flag = (i + 1) % list.size();

list.get((i+1)%list.size()).signal();

}else{

try {

list.get(i%list.size()).await();

} catch (InterruptedException e) {

e.printStackTrace();

}

}

}finally{

lock.unlock();

}

}

}



private static class Task implements Runnable {

private final int i;



public Task(int i) {

this.i = i;

}



@Override

public void run() {

work(i);

}



}



/**

* @param args

*/

public static void main(String[] args) {

for (int i = 0; i < list.size(); i++)

exec.execute(new Task(i));

}



}

解法二：
http://blog.csdn.net/lyx2007825/article/details/7767408