给定一个链表的头指针，要求只遍历一次，将单链表中的元素顺序反转过来

经典题目，代码如下：

 1 #include <iostream>
 2 #include <vector>
 3 using namespace std;
 4 struct ListNode {
 5     int m_iData;
 6     ListNode* m_pNext;
 7 };
 8 ListNode* Reverse(ListNode* in_pListNode) {
 9     if(in_pListNode == NULL) {
10         return NULL;
11     }else if(in_pListNode->m_pNext == NULL) {
12         return in_pListNode;
13     }
14     ListNode* t_pFirst = in_pListNode->m_pNext;
15     ListNode* t_pSecond = in_pListNode->m_pNext->m_pNext;
16     in_pListNode->m_pNext = NULL;
17     while(t_pFirst != NULL && t_pSecond != NULL) {
18         t_pFirst->m_pNext = in_pListNode;
19         in_pListNode = t_pFirst;
20         t_pFirst = t_pSecond;
21         t_pSecond = t_pSecond->m_pNext;
22     }
23     t_pFirst->m_pNext = in_pListNode;
24     return t_pFirst;
25 }
26
27 int main() {
28     ListNode* t_iListNode = new ListNode();
29     ListNode* t_iListNodeHead = t_iListNode;
30     for(int i = 1; i < 10; i ++) {
31         t_iListNode->m_iData = i;
32         ListNode* t_iListNodeNext = new ListNode();
33         t_iListNode->m_pNext = t_iListNodeNext;
34         t_iListNode = t_iListNodeNext;
35     }
36     t_iListNode->m_iData = 10;
37     ListNode* t_iTempListNode =t_iListNodeHead;
38     cout << "链表翻转前：" ;
39     while(t_iTempListNode != NULL) {
40         cout << t_iTempListNode->m_iData << " ";
41         t_iTempListNode = t_iTempListNode->m_pNext;
42     }
43     cout << endl;
44
45     t_iTempListNode = Reverse(t_iListNodeHead);
46
47     cout << "链表翻转后：" ;
48     while(t_iTempListNode != NULL) {
49         cout << t_iTempListNode->m_iData << " ";
50         t_iTempListNode = t_iTempListNode->m_pNext;
51     }
52     cout << endl;
53       system("pause");
54       return 0;
55
56
57 }