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。