17 合并两个排序的链表

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

C++：迭代

 1 /*
 2 struct ListNode {
 3     int val;
 4     struct ListNode *next;
 5     ListNode(int x) :
 6             val(x), next(NULL) {
 7     }
 8 };*/
 9 class Solution {
10 public:
11     ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
12     {
13         if (pHead1 == NULL)
14             return pHead2 ;
15         if (pHead2 == NULL)
16             return pHead1 ;
17         ListNode* pHead = NULL ;
18         if (pHead1->val <= pHead2->val){
19             pHead = pHead1 ;
20             pHead1 = pHead1->next ;
21         }else{
22             pHead = pHead2 ;
23             pHead2 = pHead2->next ;
24         }
25         ListNode* p = pHead ;
26         while(pHead1 != NULL && pHead2 != NULL){
27             if (pHead1->val <= pHead2->val){
28                 p->next = pHead1 ;
29                 pHead1 = pHead1->next ;
30             }else{
31                 p->next = pHead2 ;
32                 pHead2 = pHead2->next ;
33             }
34             p = p->next ;
35         }
36         if (pHead1 != NULL)  p->next = pHead1 ;
37         if (pHead2 != NULL)  p->next = pHead2 ;
38         return pHead ;
39     }
40 };

C++：递归

 1 /*
 2 struct ListNode {
 3     int val;
 4     struct ListNode *next;
 5     ListNode(int x) :
 6             val(x), next(NULL) {
 7     }
 8 };*/
 9 class Solution {
10 public:
11     ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
12     {
13         if (pHead1 == NULL)
14             return pHead2 ;
15         if (pHead2 == NULL)
16             return pHead1 ;
17         if (pHead1->val <= pHead2->val){
18             pHead1->next = Merge(pHead1->next , pHead2) ;
19             return pHead1 ;
20         }else{
21             pHead2->next = Merge(pHead1 , pHead2->next) ;
22             return pHead2 ;
23         }
24     }
25 };