算法：合并排序与链表缺少的项目

Question

I am trying to sort my linked list with merge sort. The list is actually sorted but it is kind of missing first item(s).

Merge sort functions:

public Node mergeSort(Node head) {
    if (head == null || head.next == null) {
        return head;
    }

    Node middle = middleElement(head);
    Node nextofMiddle = middle.next;
    middle.next = null;
    return merge(mergeSort(head), mergeSort(nextofMiddle));
}

public Node merge(Node left, Node right) {

    Node temp = new Node();
    Node newHead = temp;
    while (left != null && right != null) {
        if (left.info <= right.info) {
            temp.next = left;
            temp = left;
            left = temp.next;
        } else {
            temp.next = right;
            temp = right;
            right = temp.next;
        }
    }
    temp.next = (left == null) ? right : left;
    return newHead;
}

public Node middleElement(Node head) {
    if (head == null) {
        return head;
    }
    Node slow = head;
    Node fast = head;
    while (fast.next != null && fast.next.next != null) {
        slow = slow.next;
        fast = fast.next.next;
    }
    return slow;
}

So I printed the list to the screen using traverse:

    public static void main(String[] args) {
    MyLinkedList mll = new MyLinkedList();

    mll.insert(3);
    mll.insert(5);
    mll.insert(9);
    mll.insert(1);
    mll.insert(8);
    mll.insert(7);
    mll.insert(2);



    mll.mergeSort(mll.head);
    mll.traverse();

}

I have result like this: 1 and 2 missing!

After checking, i noticed that the "tail" of the linked list value is still 2. I don't know why can someone help?. I'm really new to programming so sorry for any inconvenience. Thank you for reading!

Answer 1

You have a couple problems that I can see. One, mentioned by @rcgldr, in the comments is that you merge method is returning the temp node newHead , but that node was no value and is not part of your graph. You need to return it's child newHead.next .

The other problem is that you never reset the head of your list. So after all the sorting, the head of mll still points to the original head, which in this case is 3 . So when you traverse the list you skip 1 and 2 . I'm not sure what your linked list class looks like but something like this should be close -- just assign the final node returned by mergeSort to the head of the list. Then when you traverse you should be starting in the right spot:

    mll.head = mll.mergeSort(mll.head);
    mll.traverse();

Answer 2

You're not far from a correct solution. I was able to get a working sort by adding 1 line and changing 3.

Your merge isn't doing what you think it is.

Allocating a false head node (what you call temp , not a good choice of name; try falseHead ) is a fine idea. Then falseHead.next is the true head. That's what you'll finally return as the sorted list. Note that initially it's null , which is what you'd expect.

What you're missing is a variable tail to reference the current last node in the merged list. Since there is initially no last node at all, this should be initialized equal to falseHead . Consequently, when you append the first element to the tail of the current result by setting tail.next , you'll also be setting falseHead.next , ie creating the head element. It all works out.

Now, what is the logic for, say, removing the current head of the right list and appending it to the merge result? It's just 3 steps:

1. Do the append operation by making tail.next the current head of right .
2. Update tail to tail.next .
3. Remove the head of right by updating right to right.next .

Of course the left side is similar.

Good luck. You're close.