【剑指offer 面试题17】合并两个排序的链表

思路：

　　比较两个链表端点值的大小，通过递归的方式排列。

 1 #include <iostream>
 2 using namespace std;
 3
 4 struct ListNode
 5 {
 6     int val;
 7     ListNode *next;
 8     ListNode(int v = 0):val(v), next(NULL){}
 9 };
10
11 ListNode *SortedListMerge(ListNode *phead1, ListNode *phead2)
12 {
13     if(phead1 == NULL)
14         return phead2;
15     else if(phead2 == NULL)
16         return phead1;
17
18     ListNode *pMergedHead = NULL;
19
20     if(phead1->val < phead2->val)
21     {
22         pMergedHead = phead1;
23         pMergedHead->next = SortedListMerge(phead1->next, phead2);
24     }
25     else
26     {
27         pMergedHead = phead2;
28         pMergedHead->next = SortedListMerge(phead1, phead2->next);
29     }
30
31     return pMergedHead;
32 }
33
34 int main()
35 {
36
37     ListNode *l1 = new ListNode(1);
38     ListNode *h1 = l1;
39     for(int i = 3; i <= 11; i += 2)
40     {
41         ListNode *temp = new ListNode(i);
42         l1->next = temp;
43         l1 = l1->next;
44     }
45
46     ListNode *l2 = new ListNode(2);
47     ListNode *h2 = l2;
48     for(int i = 4; i <= 12; i += 2)
49     {
50         ListNode *temp = new ListNode(i);
51         l2->next = temp;
52         l2 = l2->next;
53     }
54
55     cout<<"List1: ";
56     ListNode *print1 = h1;
57     while(print1 != NULL)
58     {
59         cout<<print1->val<<" ";
60         print1 = print1->next;
61     }
62     cout<<endl;
63
64     cout<<"List2: ";
65     ListNode *print2 = h2;
66     while(print2 != NULL)
67     {
68         cout<<print2->val<<" ";
69         print2 = print2->next;
70     }
71     cout<<endl;
72
73     ListNode *mergeList = SortedListMerge(h1, h2);
74
75     cout<<"ListMerge: ";
76     ListNode *printlist = mergeList;
77     while(printlist != NULL)
78     {
79         cout<<printlist->val<<" ";
80         printlist = printlist->next;
81     }
82     cout<<endl;
83
84
85 }

测试结果：

List1: 1 3 5 7 9 11
List2: 2 4 6 8 10 12
ListMerge: 1 2 3 4 5 6 7 8 9 10 11 12