【LeetCode】021. 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.

Example:

Input: 1->2->4, 1->3->4
Output: 1->1->2->3->4->4

题解:

  简单的链表遍历,还可用递归做。

Solution 1

 1 v/**
 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 dummy(-1);
13         ListNode* cur = &dummy;
14         
15         while(l1 && l2) {
16             if (l1->val < l2->val) {
17                 cur->next = l1;
18                 l1 = l1->next;
19             } else {
20                 cur->next = l2;
21                 l2 = l2->next;
22             }
23             cur = cur->next;
24         }
25         
26         cur->next = l1 ? l1 : l2;
27         
28         return dummy.next;
29     }
30 };

 

  递归方法

Solution 2 

 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) return l2;
13         if (!l2) return l1;
14         
15         if (l1->val < l2->val) {
16             l1->next = mergeTwoLists(l1->next, l2);
17             return l1;
18         } else {
19             l2->next = mergeTwoLists(l1, l2->next);
20             return l2;
21         }
22     }
23 };

 

posted @ 2018-03-25 20:41  Vincent丶丶  阅读(163)  评论(0编辑  收藏  举报