将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。

示例 1：
输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]

示例 2：
输入：l1 = [], l2 = []
输出：[]

示例 3：
输入：l1 = [], l2 = [0]
输出：[0]

我的解法：

没有用递归，用的笨办法，创建了一个新的链表记录合并结果

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode mergeTwoLists(ListNode list1, ListNode list2) {ListNode dummy = null, end = null;if (list1 == null && list2 == null){return dummy;} else{dummy = new ListNode(0);end = dummy;}// if (list1 == null && list2 != null) return list2;// if (list1 != null && list2 == null) return list1;while (list1 != null && list2 != null){if (list1.val <= list2.val){end.next = list1;end = end.next;list1 = list1.next;}else{end.next = list2;end = end.next;list2 = list2.next;}}if (list1 == null) end.next = list2;if (list2 == null) end.next = list1;return dummy.next;}
}// 没有用递归，用的笨办法，创建了一个新的链表记录合并结果

递归解法：

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {public ListNode mergeTwoLists(ListNode l1, ListNode l2) {if (l1 == null) {return l2;} else if (l2 == null) {return l1;} else if (l1.val < l2.val) {l1.next = mergeTwoLists(l1.next, l2);return l1;} else {l2.next = mergeTwoLists(l1, l2.next);return l2;}}
}