java - 如何使用多线程查找数字范围的总和？

Question

系统应该计算一个数字范围内的 x 个线程数的总和，其中数字范围和 x 个线程数由用户输入。

如果要在 2 个线程中计算范围是 1-4 的示例:

Thread 1: 1 + 2 = 3
Thread 2: 3 + 4 = 7

Total = 10

MyThread 类只是我为更好地理解线程的工作原理而编写的示例，而不是解决方案的实际尝试。

我的线程类:

class MyThread implements Runnable {
    private String name;

    MyThread(String thread) {
        name = thread;
        System.out.println("Start: " + name);
        try { Thread.sleep(500); } catch (Exception e) { System.out.println(e); }
    }

    //countdown from 5 to 1 for each thread
    @Override
    public void run() {
        for (int i=5; i>0; i--){
            System.out.println(name + ": " + i);
            try { Thread.sleep(500); } catch (Exception e) { System.out.println(e); }
        }
        System.out.println(name + " Exiting");
    }
}

主类:

public class SimpleThreadExample {
    public static void main (String[] args) throws InterruptedException {
        Thread t = new Thread();

        //get number of threads from user
        System.out.print("Enter the number of threads: ");
        Scanner input = new Scanner(System.in);
        int number = input.nextInt();

        //create the threads
        for (int i=1; i<=number; i++) {
            t = new Thread(new MyThread("Thread " + i));
            t.start();
        }

        t.join();
        System.out.println("End");
    }
}

我目前不知道应该如何在不同的线程中显示和添加数字。任何建议将不胜感激。

Answer 1

一种简单的 Java-8 方法是使用并行流。

int[] numbers = {1,2,3,4,5,6,7,8,9,10};

int sum = Arrays.stream (numbers).parallel()
                        .sum ();

如果您需要通过手动创建线程来完成。您可以围绕以下几行做一些事情。

    CountDownLatch countDownLatch = new CountDownLatch (2);

    SummationTask task1 = new SummationTask (countDownLatch,Arrays.copyOfRange (numbers, 0, numbers.length / 2));
    SummationTask task2 = new SummationTask (countDownLatch,Arrays.copyOfRange (numbers, (numbers.length / 2), numbers.length));
    Thread thread = new Thread (task1);
    Thread thread2 = new Thread (task2);
    thread.start ();
    thread2.start ();
    try {
        countDownLatch.await ();
        int totalSum = task1.sum + task2.sum;
        System.out.println ("sum => " + totalSum);
    } catch (InterruptedException e) {
        e.printStackTrace ();
    }

在 SummationTask 中，添加拆分数组元素。

class SummationTask implements Runnable {

    CountDownLatch countDownLatch;
    int[] arr;
    int sum;

    public SummationTask(CountDownLatch countDownLatch, int[] arr) {
        this.countDownLatch = countDownLatch;
        this.arr = arr;
    }

    @Override
    public void run() {
        int sum = Arrays.stream (arr)
                .sum ();
        this.sum = sum;
        countDownLatch.countDown ();
    }
}

PS:上面的解决方案创建了 2 个线程，您可以通过使用所需大小的 ThreadPool(按间隔拆分数组)以及 Callable 和 Future 来改进它.如果您使用 Future

，则可以摆脱 CountDownLatch