为什么超出了内存限制，我该如何避免呢? [英] Why memory limit exceeded and how can i avoid this?

问题描述

 import java.util.*;

 public class AlmostPerfect {

 public static void main(String[] args) {
 Scanner x = new Scanner(System.in);
 while(x.hasNext()){
    int n = x.nextInt();
    int sum = recursion(n,n-1,0);
    if (sum == n) {
        System.out.println(n + " perfect");
    } else if ((n - sum) <= 2) {
        System.out.println(n + " almost perfect");
    } else {
        System.out.println(n + " not perfect");
    }
}
}

public static int recursion(int n, int x,int sum) {

    if(x==0){
        return sum;
    }
    else if (n % x == 0) {
        sum+=x;
        return recursion(n,x-1,sum);
    }
    else{
        return recursion(n,x-1,sum);
    }

}

}

我想基本上找出解决方案有什么问题.存在解决方案，但是我无法理解超出内存限制的属性. 问题链接: https://open.kattis.com/problems/almostperfect 谢谢.

I want to basically find whats wrong with my solution... There exists a solution for this but I can't understand the memory limit exceeded property. Problem link: https://open.kattis.com/problems/almostperfect Thanks.

推荐答案

如果必须使用递归解决方案，则可以将递归的深度限制为 避免堆栈溢出.
您可以在连续的时间间隔(一次为一个时间间隔)上运行递归解决方案:

If you must of prefer a recursive solution, you can limit the depth of the recursion, to avoid stack overflow.
You can do it be running the recursive solution on continuous intervals, one interval at time :

import java.util.stream.IntStream;

public class AlmostPerfect {

    //defines the recursive iteration depth. increase it and you run into
    //stack over flow
    private static int RECURSION_DEPTH =5000;

    public static void main(String[] args) {

        //run comparison test between recursive and non recursive solutions
        IntStream.range(1,200000).forEach(i ->{

            double sumRecursive = recursionWithLimitedDepth(i);
            double sumIterative = iterative(i);

            if((sumRecursive != sumIterative)) {
                System.out.println("for "+ i +" "+ sumRecursive + "<>" +sumIterative );
                return;
            }

            if((i%20000) ==0) {
                System.out.println("20,000 numbers successfully checked"  );
            }
        });

        System.out.println("Test finished");
    }

    //helper method for recursive solution
    public static double recursionWithLimitedDepth(int n) {

        double sum = 0;
        int rangeStart = n-1;
        int rangeEnd = rangeStart - RECURSION_DEPTH;

        while (rangeStart > 0) {
            sum += recursionWithLimitedDepth(n, rangeStart, rangeEnd, 0);
            rangeStart = (rangeEnd - 1) >= 0 ? rangeEnd - 1 : 0;
            rangeEnd = (rangeStart - RECURSION_DEPTH) >= 0 ?  rangeStart - RECURSION_DEPTH :0;
        }

        return sum;
    }

    //run recursive solution on a limited range defined by rangeStart, rangeEnd
    public static double recursionWithLimitedDepth(int numberToTest, int rangeStart,
                                                            int rangeEnd , double sum) {
        if(rangeStart == 0){
            return sum;
        }
        else if ((numberToTest % rangeStart) == 0) {
            sum+=rangeStart;
        }

        if(rangeStart == rangeEnd){
            return sum;
        }

        return recursionWithLimitedDepth(numberToTest,rangeStart-1, rangeEnd, sum);
    }

    //simple iterative to test against
    public static double iterative(int n) {

        double sum = 0;

        for(int x = n-1; x > 0; x--) {
            if((n%x) == 0) {
                sum += x;
            }
        }

        return sum;
    }
}

请注意，sum是double，以避免Integer溢出(已通过Integer.MAX_VALUE测试).

Note that sum is double to avoid Integer overflow (tested with Integer.MAX_VALUE).

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