一道很经典的多线程面试题

最新推荐文章于 2024-07-07 21:32:49 发布

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package com.lyzx.restdy.callable;

import java.util.concurrent.Callable;
import java.util.concurrent.CountDownLatch;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.ExecutorService;
import java.util.concurrent.Executors;
import java.util.concurrent.Future;

/**
* 关于多线程的一道面试题
* 对于一个大数组求和(为模拟真实的情况没加一次等待20ms),
* 要求使用多线程实现
* 大致思路是把一个大数组分开，每个线程值计算其中一部分的和
* 最后把结果合并
*/
public class T {

    public static void main(String[] args) {
        //数组长度
        int itemCount =2000;
        int[] arr = new int[itemCount];
        for(int i=0;i<itemCount;i++){
            arr[i]=i;
        }

        //线程数，使用线程池管理线程
        int threadCount = 10;
        ExecutorService es = Executors.newFixedThreadPool(threadCount);

        //使用计数器阻塞主线程，等待所有的数组求和完毕后放开主线程
        CountDownLatch latch = new CountDownLatch(threadCount);

        //分配每个线程计算一部分数组最后求和
        int perItem=itemCount/threadCount;
        Future<String>[] allFuture = new Future[perItem];

        for(int i=0;i<threadCount;i++){
            Future<String> f = es.submit(new Z(arr,i*perItem,perItem*(i+1),latch));
            allFuture[i]=f;
        }
        es.shutdown();
        int finalCount = 0;

        for(int i=0;i<threadCount;i++){
            try {
                String value = allFuture[i].get();
                finalCount += Integer.parseInt(value);
            } catch (InterruptedException e) {
                e.printStackTrace();
            } catch (ExecutionException e) {
                e.printStackTrace();
            }
        }

        //阻塞主线程等待计算数组和的线程执行完毕
        try {
            latch.await();
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println(finalCount);
    }
}

class Z implements Callable<String>{
    //[start,end)
    private int start =-1;
    private int end =-1;
    private int[] arr;
    private int count = 0;
    private CountDownLatch latch ;


    public Z(int[] arr,int start,int end,CountDownLatch latch){
        if(arr == null || start <0 || end <=0 || start > end || latch == null){
            throw new RuntimeException("param is Error");
        }
        this.arr = arr;
        this.start=start;
        this.end=end;
        this.latch = latch;
    }

    /**
     * 对于arr数组的start到end项累加
     */
    @Override
    public String call() throws Exception {
        for(int i=start;i<end;i++){
            //每相加一次就等待20毫秒，没有实际意义，只为了模拟实际的生成环境，可能出现的阻塞情况
            Thread.sleep(20);
            count += i;
        }
        latch.countDown();
        return String.valueOf(count);
    }

}