21. Merge Two Sorted Lists

题目：将两个有序链表合并为一个新链表。该新链表必须是之前两个链表的节点合并而成。

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

zwangbo的做法

采用递归的方法，思路简洁，实现简单，每次比较返回较小值

public class Solution {
    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1 == null){
            return l2;
        }
        if(l2 == null){
            return l1;
        }
        
        ListNode mergeHead;
        if(l1.val < l2.val){
            mergeHead = l1;
            mergeHead.next = mergeTwoLists(l1.next, l2);
        }
        else{
            mergeHead = l2;
            mergeHead.next = mergeTwoLists(l1, l2.next);
        }
        return mergeHead;
    }
}

我的做法（实现复杂，提交了三次才通过）

class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode maxPointer = l1;
ListNode minPointer = l2;
ListNode maxHead = l1;
ListNode minHead = l2;

if (minPointer == null)
return maxPointer;
if (maxPointer == null)
return minPointer;

if (l1.val <= l2.val) {
minPointer = l1;
maxPointer = l2;
maxHead = l2;
minHead = l1;
}

ListNode head = minHead;

while (minPointer != null && maxPointer != null) {
while ((minPointer.next != null) && (minPointer.next.val > maxPointer.val)) {
maxHead = maxPointer.next;
minHead = minPointer.next;

minPointer.next = maxPointer;
maxPointer.next = minHead;
minPointer = maxPointer;
if (maxHead != null)
maxPointer = maxHead;
else
return head;
}
if (minPointer.next != null)
minPointer = minPointer.next;
else
while (maxPointer != null) {
minPointer.next = maxPointer;
minPointer = minPointer.next;
maxPointer = maxPointer.next;
}

}

return head;
}
}