order_print_num
问题分析:首先我们需要看清楚题目,多线程、顺序打印、然后就是几个线程交替顺序打印;
首先看下输出结果:
thread-0:0
thread-1:1
thread-2:2
thread-0:3
thread-1:4
thread-2:5
thread-0:6
thread-1:7
thread-2:8
thread-0:9
thread-1:10
thread-2:11
thread-0:12
thread-1:13
thread-2:14
thread-0:15
thread-1:16
thread-2:17
thread-0:18
thread-1:19
thread-2:20
thread-0:21
thread-1:22
thread-2:23
thread-0:24
thread-1:25
thread-2:26
thread-0:27
thread-1:28
thread-2:29
thread-0:30
thread-1:31
thread-2:32
thread-0:33
thread-1:34
thread-2:35
thread-0:36
thread-1:37
thread-2:38
thread-0:39
thread-1:40
thread-2:41
thread-0:42
thread-1:43
thread-2:44
thread-0:45
thread-1:46
thread-2:47
thread-0:48
thread-1:49
thread-2:50
thread-0:51
thread-1:52
thread-2:53
thread-0:54
thread-1:55
thread-2:56
thread-0:57
thread-1:58
thread-2:59
thread-0:60
thread-1:61
thread-2:62
thread-0:63
thread-1:64
thread-2:65
thread-0:66
thread-1:67
thread-2:68
thread-0:69
thread-1:70
thread-2:71
thread-0:72
thread-1:73
thread-2:74
thread-0:75
thread-1:76
thread-2:77
thread-0:78
thread-1:79
thread-2:80
thread-0:81
thread-1:82
thread-2:83
thread-0:84
thread-1:85
thread-2:86
thread-0:87
thread-1:88
thread-2:89
thread-0:90
thread-1:91
thread-2:92
thread-0:93
thread-1:94
thread-2:95
thread-0:96
thread-1:97
thread-2:98
thread-0:99
thread-1:100
其实这个题目的重点就是怎么实现多个线程交替顺序打印,想一想,这里我们需要用到的就是线程间的通信,然后怎么通信呢?我们可以通过 synchronized + wait + notifyAll,或者 ReentrantLock + Condition + await + signalAll;
知道了线程间怎么通信,然后就是在每个线程内部怎么实现了,是不是多个线程共用一把锁,然后自然数0-100是公共资源,让3个线程去消费;
我们可以每个线程加上一个标号:0,1,2,来表示具体是哪个线程;通过一个计数器对3进行求余,余数和具体的线程标号去比较,只有当余数和线程标号相等的时候才进行打印(不等加入等待队列),打印完计数器进行自增;然后唤醒等待队列里面线程去获取锁,获取锁成功则继续执行以上步骤,否则加入锁的同步队列中,等待上一个线程唤醒;
用 synchronized 锁实现如下:
class PrintThread1 implements Runnable {
private static final Object LOCK = new Object();
private static int count = 0; // 计数,同时确定线程是否要加入等待队列,还是可以直接去资源队列里面去获取数据进行打印
private LinkedList<Integer> queue;
private Integer threadNo;
public PrintThread1(LinkedList<Integer> queue, Integer threadNo) {
this.queue = queue;
this.threadNo = threadNo;
}
@Override
public void run() {
while (true) {
synchronized (LOCK) {
while (count % 3 != this.threadNo) {
if (count >= 101) {
break;
}
try {
LOCK.wait();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
if (count >= 101) {
break;
}
Integer val = this.queue.poll();
System.out.println("thread-" + this.threadNo + ":" + val);
count++;
LOCK.notifyAll();
}
}
}
}
用 ReentrantLock 锁实现如下:
class PrintThread2 implements Runnable {
private static final ReentrantLock lock = new ReentrantLock();
private static final Condition c = lock.newCondition();
private static int count = 0; //作为计数,同时也作为资源;因为这道题目是自然数作为资源,所以正好可以公用;
private Integer threadNo;
public PrintThread2(Integer threadNo) {
this.threadNo = threadNo;
}
@Override
public void run() {
while (true) {
try {
lock.lock();
while (count % 3 != this.threadNo) {
if (count >= 101) {
break;
}
try {
c.await();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
if (count >= 101) {
break;
}
System.out.println("thread-" + this.threadNo + ":" + count);
count++;
c.signalAll();
} finally {
lock.unlock();
}
}
}
}
具体代码实现:
参见:code
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