java - java中如何在一个线程完成时结束其他线程的处理

Question

我有多个线程正在执行搜索。我希望当一个线程完成搜索并找到解决方案时，所有其他线程都会停止运行。这是我到目前为止所拥有的

import java.util.Scanner;


class NewThread extends Thread  
{ 

    int n = 4;

    NewThread(String threadname, ThreadGroup tgob, int n) 
    { 
        super(tgob, threadname);
        this.n = n;
        start(); 
    } 
public void run() 
    { 

        System.out.println("Thread running");

                   long timestamp1 = System.currentTimeMillis();

                   System.out.println("Solution to "+ n +" queens using hill climbing search:");

                   HillClimbingSearch hcs = new HillClimbingSearch(n);

                   hcs.runSearch();

                   if (hcs.getFinalSolution() != null)
                   hcs.printState(hcs.getFinalSolution());

                   //Printing the solution          
                   long timestamp2 = System.currentTimeMillis();
                   long timeDiff = timestamp2 - timestamp1;
                   System.out.println("Execution Time: "+timeDiff+" ms");

                   System.out.println(Thread.currentThread().getName() + 
                   " finished executing"); 
    } 
}  

public class Main extends Thread{

    public static void main(String[] args) {


            int n = 0; 
            try (Scanner s=new Scanner(System.in)) {
                while (true){
                    System.out.println("Enter the number of Queens :");
                    n = s.nextInt();
                    if ( n == 2 || n ==3) {
                        System.out.println("No Solution possible for "+ n +" Queens. Please enter another number");
                    }
                    else
                        break;
                }
            }

            // creating the thread group 
            ThreadGroup gfg = new ThreadGroup("parent thread group"); 

            NewThread t1 = new NewThread("one", gfg, n); 
            System.out.println("Starting one"); 
            NewThread t2 = new NewThread("two", gfg, n); 
            System.out.println("Starting two"); 
            NewThread t3 = new NewThread("three", gfg, n); 
            System.out.println("Starting three"); 

            boolean keepRunning = true;

            while(keepRunning){
                if (t1.isAlive() && t2.isAlive() && t3.isAlive()){
                    continue;
                } else {
                    t1.interrupt();
                    t2.interrupt();
                    t3.interrupt();
                    keepRunning = false;
                }
            }

            // checking the number of active thread 
            System.out.println("number of active thread: "
                               + gfg.activeCount()); 

        }
}

这会编译并打印一个解决方案，但它会从仍在运行的每个线程打印多个解决方案。

我的输出看起来像这样

Enter the number of Queens :
5
Starting one
Thread running
Starting two
Starting three
Thread running
Thread running
Solution to 5 queens using hill climbing search:
Solution to 5 queens using hill climbing search:
Solution to 5 queens using hill climbing search:


0 1 0 0 0 
0 0 0 1 0 
1 0 0 0 0 
0 0 1 0 0 
0 0 0 0 1 
0 1 0 0 0 
0 0 0 1 0 
1 0 0 0 0 
0 0 1 0 0 
0 0 0 0 1 
0 1 0 0 0 
0 0 0 1 0 
1 0 0 0 0 
0 0 1 0 0 
0 0 0 0 1 
Execution Time: 81 ms
Execution Time: 72 ms
Execution Time: 98 ms
one finished executing
two finished executing
three finished executing
number of active thread: 0

感谢您的帮助或建议。

Answer 1

这是给您的建议。

首先，您需要一个方法来遍历线程集合并向每个线程发出停止信号。为此，您必须将线程集合 gfg 设置为实例变量，而不是 main 方法的本地变量。

您编写的用于关闭线程的方法应该迭代 gfg 中的每个 NewThread 并在其中设置一个 boolean 值作为退出时间的信号。

NewThread 类的 run() 方法应定期查看此 boolean 值。多久一次取决于你。如果 run() 发现 boolean 值已设置，则通过安静的分支退出。

最后一件事，当 NewThread 完成任务时，它应该调用您的方法来关闭线程。