启动3个线程打印递增的数字, 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, 直到打印到75. 程序的输出结果应该为:
线程1: 1
线程1: 2
线程1: 3
线程1: 4
线程1: 5
线程2: 6
线程2: 7
线程2: 8
线程2: 9
线程2: 10
...
线程3: 71
线程3: 72
线程3: 73
线程3: 74
线程3: 75
public class PrintNoDemo { private static int printNO = 1; private static int printThread = 1; private static Condition condition1; private static Condition condition2; private static Condition condition3; private static ReentrantLock lock = new ReentrantLock(); public static void main(String[] args){ condition1 = lock.newCondition(); condition2 = lock.newCondition(); condition3 = lock.newCondition(); for(int i=0; i<3; i++){ Thread thread = new Thread(new Runnable(){ @Override public void run() { while(printNO <= 75){ printNo(); } } },"线程"+(i+1)); thread.start(); } } public static void printNo() { try { lock.lock(); String threadName = Thread.currentThread().getName(); //判断线程，选择等待Condition if(!threadName.equals("线程"+printThread)){ if(threadName.equals("线程1")){ condition1.await(); } else if(threadName.equals("线程2")){ condition2.await(); } else if(threadName.equals("线程3")){ condition3.await(); } } //处理打印 for (int i = 0; i < 5; i++) { if(printNO > 75){ break; } System.out.println(Thread.currentThread().getName() +":"+ printNO++); } //获取下一个打印线程 printThread = printThread%3 +1; if(threadName.equals("线程1")){ condition2.signal(); }else if(threadName.equals("线程2")){ condition3.signal(); }else if(threadName.equals("线程3")){ condition1.signal(); } // lock.notifyAll(); } catch (InterruptedException e) { e.printStackTrace(); } finally { lock.unlock(); } }} 网友实现方法
class NumberPrinter extends Thread { static int c = 0; static int state = 0; private int id; @Override public synchronized void run() { while (state < 15) { if (state % 3 == id) { for (int j = 0; j < 5; j++) { c++; System.out.format("Thread %d: %d %n", id, c); } state++;; } } } public NumberPrinter(int id) { this.id = id; } public static void main(String[] args) { for (int i = 0; i < 3; i++) { new NumberPrinter(i).start(); } }}