我的合并排序代码有什么问题？

Question

I'm having trouble with Merge Sort in my computers class. I keep getting either errors or the original ArrayList returned.

I believe that Merge Sort involves splitting an array(list) in half recursively until there is only one element left, then, working from those individual elements, merging them in sorted order. This continues until the array(list) has been sorted. As for the actual sorting part, I am trying to insert into a new ArrayList the higher value between the two halves until they are both empty, in which case the filled ArrayList is now sorted.

Here's my current code:

public static ArrayList<Integer> mergesort(ArrayList<Integer> arr) {

    // Base case: if size of ArrayList is 1, return it.
    if (arr.size() < 2) {
        return arr;
    }

    // Else: Find the middle index.
    int middle = (arr.size() - 1) / 2;

    // Split into left and right halves.
    ArrayList<Integer> leftHalf = new ArrayList<Integer>();
    for (int i = 0; i < middle; i++)
        leftHalf.add(arr.get(i));

    ArrayList<Integer> rightHalf = new ArrayList<Integer>();
    for (int j = middle; j < arr.size(); j++)
        rightHalf.add(arr.get(j));

    // Recurse using the halves.
    mergesort(leftHalf);
    mergesort(rightHalf);

    // Sort and merge the two halves.
    return merge(leftHalf, rightHalf, arr);
}

// Merge two halves and sort them, and put the sorted values into the ArrayList sorted.
public static ArrayList<Integer> merge(ArrayList<Integer> arr1, ArrayList<Integer> arr2, ArrayList<Integer> sorted) {

    // While the ArrayLists are not empty, add sorted elements to ArrayList sorted.
    while (arr1.size() > 0 && arr2.size() > 0) {

        // If the first element in A is greater than the first in B, remove A and add to ArrayList sorted.
        if (arr1.get(0) >= arr2.get(0)) {
            sorted.add(arr1.get(0));
            arr1.remove(0);
        }
        // Else, remove from B and add to ArrayList sorted.
        else {
            sorted.add(arr2.get(0));
            arr2.remove(0);
        }
    }

    // If there're still elements in A due to arr being odd
    // add them to C since they will be the largest.
    if (arr1.size() > 0)
        sorted.add(arr1.get(0));

    return sorted;
}

I would appreciate any help, but please don't give me a full implementation of the Merge Sort, since I want to actually learn how to do this for the future.

Answer 1

Two immediate problems I see.

First, you have infinite recursion. Say you pass in a 4-value list and print your list size and middle values, you'll get:

size 4, middle 1 size 3, middle 1 size 2, middle 0 size 2, middle 0 size 2, middle 0 ... to infinity

Remember that Java is aleardy doing integer division in your code.

Second, you're passing the arr parameter from mergesort to merge , nothing is ever getting removed from it. So even if you made it past the infinite recursion you'd still end up with a larger list in the end, your original list plus anything you've .add() ed to it

Answer 2

As mentioned by MarquisDeMizzle, so that middle == 1 when arr.size() == 1.

    int middle = arr.size() / 2;

After main merge loop copy any remaining elements, and since you don't know if arr1 or arr2 go empty first, check both:

    // copy any remaining elements
    while (arr1.size() > 0)
        sorted.add(arr1.get(0));

    while (arr2.size() > 0)
        sorted.add(arr2.get(0));

If interested, you might want to learn bottom up merge sort. It skips all the recursion and just starts off assuming an array with n elements is n runs of size 1, just using indexing to do the merge back and forth between the original and a temporary array.