LeetCode：Merge Two Sorted Lists

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.

简单题，只要对两个链表中的元素进行比较，然后移动即可，只要对链表的增删操作熟悉，几分钟就可以写出来，代码如下：

 1 struct ListNode {
 2     int val;
 3     ListNode *next;
 4     ListNode(int x):val(x), next(NULL) {}
 5 };
 6
 7 ListNode *GetLists(int n)    //得到一个列表
 8 {
 9     ListNode *l = new ListNode(0);
10     ListNode *pre = l;
11     int val;
12     for (int i = 0; i < n; i ++) {
13         cin >> val;
14         ListNode *newNode = new ListNode(val);
15         pre->next = newNode;
16         pre = pre->next;
17     }
18     return l->next;
19 }
20
21 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2)
22 {
23     assert (NULL != l1 && NULL != l2);
24     if (NULL == l1 && NULL == l2)
25         return NULL;
26     if (NULL == l1 && NULL != l2) // !!要记得处理一个为空，另一个不为空的情况
27         return l2;
28     if (NULL != l1 && NULL == l2)
29         return l1;
30
31     ListNode *temp = new ListNode(0);
32     temp->next = l1;
33     ListNode *pre = temp;
34
35     while(NULL != l1 && NULL != l2) {
36         if (l1->val > l2->val) { //从小到大排列
37             ListNode *next = l2->next;
38             l2->next = pre->next;
39             pre->next = l2;
40             l2 = next;
41         }
42         else {
43             l1 = l1->next;
44         }
45         pre = pre->next;
46     }
47     if (NULL != l2) {
48         pre->next = l2;
49     }
50     return temp->next;
51 }

这其中要注意一点，即要记得处理一个链表为空，另一个不为空的情况，如{}, {0} -- > {0},当然上面的写法多少啰嗦了一些，可以简写。