LeetCode OJ:Merge Two Sorted Lists(合并两个链表)

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.

注意题目要求合并的时候不能新建节点,直接使用原来的节点,比较简单,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         if(l1 == NULL) return l2;
13         if(l2 == NULL) return l1;
14         ListNode * root = new ListNode(-1);
15         ListNode * helper = root;
16         while(l1!=NULL && l2!=NULL){
17             if(l1->val <= l2->val)
18                 root->next = l1, l1=l1->next;
19             else if(l1->val > l2->val)
20                 root->next = l2, l2=l2->next;
21             root = root->next;
22         }
23         while(l1!=NULL){
24             root->next = l1;
25             l1 = l1->next;
26             root = root->next; 
27         }
28         while(l2!=NULL){
29             root->next = l2;
30             l2 = l2->next;
31             root = root->next; 
32         }
33         return helper->next;
34     }
35 };

 

 1 public class Solution {
 2     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
 3         ListNode helper = new ListNode(0);
 4         ListNode ret = helper;
 5         if(l1 == null) return l2;
 6         if(l2 == null) return l1;
 7         while(l1 != null && l2 != null){
 8             if(l1.val < l2.val){
 9                 helper.next = l1;
10                 l1 = l1.next;
11             }else{
12                 helper.next = l2;
13                 l2 = l2.next;
14             }
15             helper = helper.next;
16         }
17         while(l1 != null){
18             helper.next = l1;
19             l1 = l1.next;
20             helper = helper.next;
21         }
22         while(l2 != null){
23             helper.next = l2;
24             l2 = l2.next;
25             helper = helper.next;
26         }
27         return ret.next;
28     }
29 }

 

posted @ 2015-11-03 21:30  eversliver  阅读(328)  评论(0编辑  收藏  举报