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 }