Java MergeSort-内存不足错误:Java堆空间 [英] Java MergeSort - Out Of Memory Error: Java Heap Space

问题描述

我正在尝试使用Java进行排序.

I'm trying to get some practice with sorting in Java.

我正在处理合并排序... Eclipse正在输出Out Of Memory Error: Java Heap space，但是我不确定如何调试它.

I'm working on the merge sort now... Eclipse is outputting Out Of Memory Error: Java Heap space, but I'm not sure how to debug that.

我觉得我的代码还可以，有什么想法吗?

I feel like my code is okay- any thoughts?

import java.util.ArrayList;
import java.util.List;
public class Sorts {
    List<Integer> initialList;

    public Sorts() {
        initialList = new ArrayList<Integer>();
        initialList.add(2);
        initialList.add(5);
        initialList.add(9);
        initialList.add(3);
        initialList.add(6);

        System.out.print("List: [");
        for (int values : initialList) {
            System.out.print(values);
        }
        System.out.println("]");

        splitList(initialList);
    }

    public List<Integer> splitList(List<Integer> splitMe)   {
        List<Integer> left = new ArrayList<Integer>();
        List<Integer> right = new ArrayList<Integer>();

        if (splitMe.size() <= 1) {
            return splitMe;
        }

        int middle = splitMe.size()/2;
        int i = 0;
        for (int x: splitMe) {
            if (i < middle) {
                left.add(x);
            }
            else {
                right.add(x);
            }
            i++;
        }
        left = splitList(left);
        right = splitList(right);

        return mergeThem(left, right);
    }

    public List<Integer> mergeThem(List<Integer> left, List<Integer> right) {
        List<Integer> sortedList = new ArrayList<Integer>();
        int x = 0;
        while (left.size() > 0 || right.size() > 0) {
            if (left.size() > 0 && right.size() > 0) {
                if (left.get(x) > right.get(x)) 
                    sortedList.add(left.get(x));
                else 
                    sortedList.add(right.get(x));
            }
            else if (left.size() > 0) {
                sortedList.add(left.get(x));
            }
            else if (right.size() > 0) {
                sortedList.add(right.get(x));
            }
        }
        return sortedList;
    }   
}

推荐答案

使用Java元素提供mergeThem方法的可能实现:

Providing a possible implementation of the mergeThem method using Java elements:

public List<Integer> mergeThem(List<Integer> left, List<Integer> right) {
    //set the sorted list
    List<Integer> sortedList = new ArrayList<Integer>();
    //getting the iterators for both lists because List#get(x) can be O(N) on LinkedList
    Iterator<Integer> itLeft = left.iterator();
    Iterator<Integer> itRight = right.iterator();
    //getting flags in order to understand if the iterator moved
    boolean leftChange = true, rightChange = true;
    //getting the current element in each list
    Integer leftElement = null, rightElement = null;
    //while there are elements in both lists
    //this while loop will stop when one of the list will be fully read
    //so the elements in the other list (let's call it X) must be inserted
    while (itLeft.hasNext() && itRight.hasNext()) {
        //if left list element was added to sortedList, its iterator must advance one step
        if (leftChange) {
            leftElement = itLeft.next();
        }
        //if right list element was added to sortedList, its iterator must advance one step
        if (rightChange) {
            rightElement = itRight.next();
        }
        //cleaning the change flags
        leftChange = false;
        rightChange = false;
        //doing the comparison in order to know which element will be inserted in sortedList
        if (leftElement <= rightElement) {
            //if leftElement is added, activate its flag
            leftChange = true;
            sortedList.add(leftElement);
        } else {
            rightChange = true;
            sortedList.add(rightElement);
        }
    }
    //this is the hardest part to understand of this implementation
    //java.util.Iterator#next gives the current element and advance the iterator on one step
    //if you do itLeft.next then you lost an element of the list, that's why we have leftElement to keep the track of the current element of left list (similar for right list)
    if (leftChange && rightElement != null) {
        sortedList.add(rightElement);
    }
    if (rightChange && leftElement != null) {
        sortedList.add(leftElement);
    }
    //in the end, you should add the elements of the X list (see last while comments).
    while (itLeft.hasNext()) {
        sortedList.add(itLeft.next());
    }
    while (itRight.hasNext()) {
        sortedList.add(itRight.next());
    }
    return sortedList;
}

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