Merge Two Sorted Lists

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if(l1 == null) return l2;
if(l2 == null) return l1;
ListNode pointer = new ListNode(0);
ListNode answer = pointer;
int i = 0;
while(l1 != null || l2 != null)
{
if(l1 == null)
{
pointer.next = l2;
pointer = pointer.next;
l2 = l2.next;
}
else if(l2 == null)
{
pointer.next = l1;
pointer = pointer.next;
l1 = l1.next;
}
else if(l1.val <= l2.val )
{
pointer.next = l1;
pointer = pointer.next;
l1 = l1.next;
}
else if(l1.val > l2.val )
{
pointer.next = l2;
pointer = pointer.next;
l2 = l2.next;
}
}
return answer.next;
}
}

上面这个是自己的solution，下面的是leetcode上别人的solution。

ListNode pointer = new ListNode(0);//ListNode pointer = new ListNode();
ListNode answer = pointer;//ListNode answer =pointer;
有一点不明白就是上面片段的语句按照注释里写怎么就不对了，求大神教学啊!ListNode(0)是初始化 pointer为{ 0 }么?

public class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0);
ListNode headcopy = head;
while(l1 != null && l2 != null){
if(l1.val < l2.val){
headcopy.next = l1;
l1 = l1.next;
}
else if(l2.val <= l1.val){
headcopy.next = l2;
l2 = l2.next;
}
headcopy = headcopy.next;
}

headcopy.next = (l1 == null ? l2 : l1);
return head.next;
}
}