【leetcode】Merge Two Sorted Lists

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.

水题不解释，一A，代码如下：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
12         ListNode* head;
13         if(l1 == NULL)
14             return l2;
15         if(l2 == NULL)
16             return l1;
17 
18         if(l1->val < l2->val)
19         {
20             head = l1;
21             l1 = l1->next;
22         }
23         else
24         {
25             head = l2;
26             l2 = l2->next;
27         }
28         ListNode* current = head;
29         while(l1 != NULL && l2 != NULL){
30             if(l1->val < l2->val){
31                 current->next = l1;
32                 l1 = l1->next;
33                 current = current->next;
34             }
35             else
36             {
37                 current->next = l2;
38                 l2 = l2->next;
39                 current = current->next;
40             }
41         }
42         if(l1 != NULL)
43             current->next = l1;
44         if(l2 != NULL)
45             current->next = l2;
46         return head;
47     }
48 };

Java版本代码：

 1 /**
 2  * Definition for singly-linked list.
 3  * public class ListNode {
 4  *     int val;
 5  *     ListNode next;
 6  *     ListNode(int x) {
 7  *         val = x;
 8  *         next = null;
 9  *     }
10  * }
11  */
12 public class Solution {
13     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
14         if(l1 == null)
15             return l2;
16         if(l2 == null)
17             return l1;
18         
19         ListNode head = null;
20         if(l1.val < l2.val)
21         {
22             head = l1;
23             l1 = l1.next;
24         }else{
25             head = l2;
26             l2 = l2.next;
27         }
28         
29         ListNode current = head;
30         while(l1 != null && l2 != null){
31             if(l1.val < l2.val){
32                 current.next = l1;
33                 l1 = l1.next;
34             }else{
35                 current.next = l2;
36                 l2 = l2.next;
37             }
38             current = current.next;
39         }
40         
41         if(l1 != null)
42             current.next = l1;
43         if(l2 != null)
44             current.next = l2;
45         
46         return head;
47     }
48 }