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.

 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         // Start typing your Java solution below
15         // DO NOT write main() function
16         ListNode root = new ListNode(-1);
17         ListNode cur1 = l1;
18         ListNode cur2 = l2;
19         ListNode cur3 = root;
20         while(cur1 != null && cur2 != null){
21             if(cur1.val < cur2.val){
22                 cur3.next = cur1;
23                 cur3 = cur1;
24                 cur1 = cur1.next;
25             }else{
26                 cur3.next = cur2;
27                 cur3 = cur2;
28                 cur2 = cur2.next;
29             }
30         }
31         if(cur1 == null){
32             cur3.next = cur2;
33         }
34         if(cur2 == null){
35             cur3.next = cur1;
36         }
37         return root.next;
38     }
39 }

第三遍：

 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         ListNode header = new ListNode(-1);
15         ListNode cur = header, cur1 = l1, cur2 = l2;
16         while(cur1 != null || cur2 != null){
17             int aa = cur1 != null ? cur1.val : Integer.MAX_VALUE;
18             int bb = cur2 != null ? cur2.val : Integer.MAX_VALUE;
19             if(aa > bb){
20                 cur.next = cur2;
21                 cur2 = cur2.next;
22                 cur = cur.next;
23             } else {
24                 cur.next = cur1;
25                 cur1 = cur1.next;
26                 cur = cur.next;
27             }
28         }
29         return header.next;
30     }
31 }