合并两个排序列表，但我的头未更新-Java

Question

I am working on "Merge two sorted linked list", but it looks my head has not been updated correctly.

To solve this problem, I tried .val, but it doesn't work correctly in while loop.

Then, the wired thing is that when I tried .next, it works. I am totally confused. I put both code below(the working one and the wrong one), so you will be able to see what I did.

Could anybody give me an explanation of why the first is not working, please?

Wrong one:

    /**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
/**
 * @param ListNode l1 is the head of the linked list
 * @param ListNode l2 is the head of the linked list
 * @return: ListNode head of linked list
 */
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    if (l1 == null)
        return l2;
    else if (l2 == null)
        return l1;

    ListNode result = new ListNode(5);
    ListNode head = result;

    while (l1 != null && l2 != null){
        if (l1.val < l2.val){
            result = l1;
            l1 = l1.next;
        }
        else{
            result = l2;
            l2 = l2.next;
        }
        result = result.next;
    }

    if (l1 == null){
        result = l2;
    }
    else{
        result = l1;
    }

    return head;

    }
}

Working one:

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
    if (l1 == null)
        return l2;
    else if (l2 == null)
        return l1;

    ListNode result = new ListNode(5);
    ListNode head = result;

    while (l1 != null && l2 != null){
        if (l1.val < l2.val){
            result.next = l1;
            l1 = l1.next;
        }
        else{
            result.next = l2;
            l2 = l2.next;
        }
        result = result.next;
    }

    if (l1 == null){
        result.next = l2;
    }
    else{
        result.next = l1;
    }

    return head.next;

}

The only difference is that I add .next in the second one.

Thank you!

Answer 1

The code uses a dummy node so that after merging, dummy_node.next will end up pointing to the merged list. This simplifies the code so it doesn't have to conditionally deal with an initially empty merged list. (In C or C++, a pointer to pointer to node could be used instead of a dummy node, but java doesn't have an equivalent.) The code starts off by setting both result and head as references to the dummy node, then advances result as the lists are merged. The working code is returning head.next, which is the original dummy_node.next as intended. The non-working code is returning head, which is a reference to the dummy node, instead of a reference to the merged list.