leetcode

题目：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.

个人思路：

1、合并两个有序链表，一般就是采用归并的方法，分别设置一个指针用于遍历链表，然后挨个比较节点大小，将小的那个节点作为新链表的下一个节点，然后让小节点的指针向后走一步，直到某个链表走完

2、当某个链表走完，则将另外一个链表的剩余部分接到新链表的后面就行了

代码：

 1 #include <stddef.h>
 2 
 3 struct ListNode
 4 {
 5     int val;
 6     ListNode *next;
 7     ListNode(int x) : val(x), next(NULL) {};
 8 };
 9 
10 class Solution
11 {
12 public:
13     ListNode* mergeTwoLists(ListNode *l1, ListNode *l2)
14     {
15         //若l1或者l2为空
16         if (!l1)
17         {
18             return l2;
19         }
20         if (!l2)
21         {
22             return l1;
23         }
24 
25         ListNode *cur_l1 = l1;
26         ListNode *cur_l2 = l2;
27         ListNode *head_new = NULL;
28         ListNode *cur_new = NULL;
29 
30         while (cur_l1 && cur_l2)
31         {
32             if (cur_l1->val <= cur_l2->val)
33             {
34                 if (!cur_new)
35                 {
36                     head_new = cur_new = cur_l1;
37                 }
38                 else
39                 {
40                     cur_new->next = cur_l1;
41                     cur_new = cur_new->next;
42                 }
43                 cur_l1 = cur_l1->next;
44             }
45             else
46             {
47                 if (!cur_new)
48                 {
49                     head_new = cur_new = cur_l2;
50                 }
51                 else
52                 {
53                     cur_new->next = cur_l2;
54                     cur_new = cur_new->next;
55                 }
56                 cur_l2 = cur_l2->next;
57             }
58         }
59 
60         if (!cur_l1)
61         {
62             while (cur_l2)
63             {
64                 cur_new->next = cur_l2;
65                 cur_new = cur_new->next;
66                 cur_l2 = cur_l2->next;
67             }
68         }
69 
70         if (!cur_l2)
71         {
72             while (cur_l1)
73             {
74                 cur_new->next = cur_l1;
75                 cur_new = cur_new->next;
76                 cur_l1 = cur_l1->next;
77             }
78         }
79 
80         return head_new;
81     }
82 };
83 
84 int main()
85 {
86     return 0;
87 }

View Code

网上基本都是这个归并的思路，属于比较基础的题目。