线程笔试题java 跪求代码思路

一个线程A以随机的间隔[0-20ms]产生k个随机数到一块内存区a (k也是随机生成的[100-200])。线程B,C每次以随机的时间间隔[0-50ms]获取内存区a的数据。B,C将获取的数据X,生成X+1。 并将X,X+1这两个数分别以随机的时间间隔[0-100ms]放入到内存区b. 线程D获取内存区b的数据并打印出来。D线程打印的数据中X与X+1必须在一起打印。当D线程打印出最后一个数字后。 输出一下打印的个数,并退出整个程序。
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不知道符不符合你的要求 public class Test { public static void main(String[] args){ BlockingQueue<Integer> que1 = new LinkedBlockingQueue<>(); BlockingQueue<MyNumber> que2 = new LinkedBlockingQueue<>(); ProducerTask A =new ProducerTask(que1); ConsumerIntegerProduceMyNumber B =new ConsumerIntegerProduceMyNumber(que1,que2); ConsumerIntegerProduceMyNumber C =new ConsumerIntegerProduceMyNumber(que1,que2); ProduceMyNumber D = new ProduceMyNumber(que2); ExecutorService executor = Executors.newFixedThreadPool(4); executor.execute(A); executor.execute(B); executor.execute(C); executor.execute(D); executor.shutdown(); try { executor.awaitTermination(Long.MAX_VALUE,TimeUnit.HOURS); } catch (InterruptedException e) { e.printStackTrace(); } } } class MyNumber{ public Integer num; public boolean isBig; public MyNumber(Integer num,boolean isBig){ this.num = num; this.isBig = isBig; } } class ProducerTask implements Runnable{ private BlockingQueue<Integer> que1; public ProducerTask(BlockingQueue<Integer> que1){ this.que1 = que1; } @Override public void run() { Random r =new Random(); int k = r.nextInt(100)+100; System.out.println(k); while(k-->0){ int time =r.nextInt(20); try { Thread.sleep(time); que1.put(r.nextInt(Integer.MAX_VALUE)); } catch (InterruptedException e) { e.printStackTrace(); } } try { que1.put(-1); que1.put(-1); } catch (InterruptedException e) { e.printStackTrace(); } } } class ConsumerIntegerProduceMyNumber implements Runnable{ private BlockingQueue<Integer> que1; private BlockingQueue<MyNumber> que2; public ConsumerIntegerProduceMyNumber(BlockingQueue<Integer> que1, BlockingQueue<MyNumber> que2) { this.que1=que1; this.que2 =que2; } @Override public void run() { int x=0 ; Random r =new Random(); while(x>=0){ int time =r.nextInt(50); try { Thread.sleep(time); x=que1.take(); if(x<0)break; int putTime1=r.nextInt(100); int putTime2 =r.nextInt(100); if(putTime1>putTime2){ Thread.sleep(putTime2); que2.put(new MyNumber(x+1,true)); Thread.sleep(putTime1-putTime2); que2.put(new MyNumber(x,false)); } else if(putTime1<putTime2){ Thread.sleep(putTime1); que2.put(new MyNumber(x,false)); Thread.sleep(putTime2-putTime1); que2.put(new MyNumber(x+1,true)); } else{ Thread.sleep(putTime1); que2.put(new MyNumber(x,false)); que2.put(new MyNumber(x+1,true)); } } catch (InterruptedException e) { e.printStackTrace(); } } try { que2.put(new MyNumber(-1,false)); } catch (InterruptedException e) { e.printStackTrace(); } } } class ProduceMyNumber implements Runnable{ private BlockingQueue<MyNumber> que2; public ProduceMyNumber(BlockingQueue<MyNumber> que2) { this.que2 = que2; } @Override public void run() { MyNumber num =new MyNumber(0,false); Set<Integer> set = new HashSet<>(); int count=0; int nThread=2; while(nThread>0){ try { num = que2.take(); if(num.num<0){ nThread--; continue; } if(num.isBig){ if(set.contains(num.num-1)){ System.out.println(num.num-1); System.out.println(num.num); count+=2; } else set.add(num.num); } else{ if(set.contains(num.num+1)){ System.out.println(num.num); System.out.println(num.num+1); count+=2; } else set.add(num.num); } } catch (InterruptedException e) { e.printStackTrace(); } } System.out.println("总计:"+count); } }
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发布于 2017-08-01 17:12

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