阿里的伯乐在线测试:机试题(生产者-消费者模式编写代码实现)

2019-10-11  本文已影响0人  Easy的幸福

使用“生产者-消费者模式”编写代码实现:线程A随机间隔(10~200ms)按顺序生成1到100的数字(共100个),
放到某个队列中.线程B、C、D即时消费这些数据,线程B消费所有被3整除的数,
线程C消费所有被5整除的数,其它的由线程D进行消费。线程BCD消费这些数据时在控制台中打印出来,
要求按顺序打印这些数据
限时40分钟,可以查API

当时我用了blockQueue去实现,没有实现出来,后来被面试官问的哑口无言,下面是自己查阅资料,并请教了我公司老大,最后给整理出来的代码。后来觉的我老大给我说的很受用,这种题一只要把题目给分析清楚了(主要考察多线程之间的相互通讯,使用lock的多路通知),一般这种题都是会有坑的,千万不要跳进去。另外,java并发包里面的的东西真的很重要。

import java.util.LinkedList;
import java.util.PrimitiveIterator;
import java.util.Queue;
import java.util.Random;
import java.util.concurrent.TimeUnit;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

public class Test {

    static class MyQueue {

        private Queue<Integer> queue = new LinkedList<>();

        private PrimitiveIterator.OfLong longs = new Random().longs(10, 200).iterator();

        private Lock lock = new ReentrantLock();

        private Condition b = lock.newCondition();
        private Condition c = lock.newCondition();
        private Condition d = lock.newCondition();

        public int b_pull() {
            lock.lock();
            try {
                try {
                    b.await();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            } finally {
                lock.unlock();
            }
            return queue.remove();
        }

        public int c_pull() {
            lock.lock();
            try {
                try {
                    c.await();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            } finally {
                lock.unlock();
            }
            return queue.remove();
        }

        public int d_pull() {
            lock.lock();
            try {
                try {
                    d.await();
                } catch (InterruptedException e) {
                    e.printStackTrace();
                }
            } finally {
                lock.unlock();
            }
            return queue.remove();
        }

        public void start() {
            new Thread(() -> {
                for (int i = 1; i <= 100; i++) {
                    queue.add(i);
                    lock.lock();
                    try {
                        if (i % 3 == 0) {
                            b.signal();
                        } else if (i % 5 == 0) {
                            c.signal();
                        } else {
                            d.signal();
                        }
                    } finally {
                        lock.unlock();
                    }
                    try {
                        TimeUnit.MILLISECONDS.sleep(longs.nextLong());
                    } catch (InterruptedException ignore) {

                    }
                }
                System.out.println("all numbers are produced.");
            }).start();
        }
    }

    public static void main(String[] args) {
        MyQueue queue = new MyQueue();
        new Thread(() -> {
            while (true) {
                System.out.println(String.format("B(mod 3) consume : %d", queue.b_pull()));
            }
        }).start();
        new Thread(() -> {
            while (true) {
                System.out.println(String.format("C(mod 5) consume : %d", queue.c_pull()));
            }
        }).start();
        new Thread(() -> {
            while (true) {
                System.out.println(String.format("D(other) consume : %d", queue.d_pull()));
            }
        }).start();
        try {
            Thread.sleep(1000L);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        queue.start();
    }
}
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