java - java中递归方法的时间限制

Question

我正在尝试用java编写一个搜索(ids)程序，如果3分钟过去并且还没有找到解决方案，该程序就会停止。我尝试使用 while 循环，但是即使找到解决方案，循环仍然继续，但时间限制仍然存在，所以我将其更改为使用 Executor 服务，但这也不断重复。我认为执行者工作不正常。我做错了什么？？

public void search(Node head) throws InterruptedException, ExecutionException{
    solution = new Stack<Node>();
    allnodes = 0;


    ExecutorService es = Executors.newSingleThreadExecutor();
    Future f = es.submit(new Runnable() {
        @Override
        public void run() {
            startTime = System.nanoTime();
            while(!Thread.interrupted()) {
                endTime = searching(start);
            }
            long time = endTime - startTime;
            System.out.printf("Finished task after %,d ns%n"+ time);
        }
    });
    try {// stops if the task completes.
        f.get(1, TimeUnit.SECONDS); 
    } catch (TimeoutException e) {
        f.cancel(true);
    }
    es.shutdown();



}



public long searching(Node head){

    solution.push(head);

        if((head.getRow() == goal.getRow()) && (head.getCol() == goal.getCol())){
            System.out.println("solution");
            endTime = System.currentTimeMillis();
            solution.push(head);
            allnodes = allnodes+1;

            return endTime; //here it returns the end time to then be printed out but the loop keeps going?



        }
        //recursively calling
        int r = head.getRow();
        int c = head.getCol();

        if((r-1 >-1 )){
            if(gr.getstate(r - 1, c) == State.Unvisited){       
                allnodes = allnodes+1;
                System.out.println("go north");
                gr.updatemap(r - 1, c);
                solution.push(head);
                searching(head.getN(head));

            }
        }
        if(r+1 < (map[1].length-1)){
            if(gr.getstate(r + 1, c) == State.Unvisited){               
                System.out.println("go south");
                allnodes = allnodes+1;
                gr.updatemap(r + 1, c);
                solution.push(head);
                head = head.getS(head);
                searching(head);    
            }
        }
        if(c+1 < map.length-1){
            if(gr.getstate(r , c +1) == State.Unvisited){
                allnodes = allnodes+1;
                System.out.println("go east");
                gr.updatemap(r , c +1);
                solution.push(head);
                head = head.getE(head);
                searching(head);
            }   
        }
        if(c -1 > -1){
            if (gr.getstate(r, c-1) == State.Unvisited){
                System.out.println("go west");
                allnodes = allnodes+1;
                gr.updatemap(r, c-1);
                solution.push(head);
                head = head.getW(head);
                searching(head);
            }   
        }


        return endTime;

}

Answer 1

关闭 ExecutorService 会阻止新任务，但不会停止现有任务。您是否尝试过使用类似的东西:

long t1 = System.getCurrentTimeMillis();
for (...) {
    // your code

    if (System.CurrentTimeMillis()-t1 > 1000*60*3) {
        break; // or even return
    }
}

您可以在任意位置放置类似的 if 语句。它会减慢你的计算速度，但会导致它按照你的预期停止。