为什么超出了内存限制，我该如何避免呢? [英] Why was the 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);
        }
    }
}

我想从根本上找到解决方案的问题.存在解决方案，但是我无法理解内存超出限制属性.

I want to basically find what's wrong with my solution... There exists a solution for this, but I can't understand the memory limit exceeded property.

问题链接: https://open.kattis.com/problems/almostperfect

推荐答案

如果必须使用递归解决方案，则可以限制递归的深度，以避免堆栈溢出.

If you must prefer a recursive solution, you can limit the depth of the recursion, to avoid stack overflow.

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:

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 overflow
    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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