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：

#include <iostream>
#include <vector>
using namespace std;
struct ListNode {
    int m_iData;
    ListNode* m_pNext;
};
ListNode* Reverse(ListNode* in_pListNode) {
    if(in_pListNode == NULL) {
        return NULL;
    }else if(in_pListNode->m_pNext == NULL) {
        return in_pListNode;
    }
    ListNode* t_pFirst = in_pListNode->m_pNext;
    ListNode* t_pSecond = in_pListNode->m_pNext->m_pNext;
    in_pListNode->m_pNext = NULL;
    while(t_pFirst != NULL && t_pSecond != NULL) {
        t_pFirst->m_pNext = in_pListNode;
        in_pListNode = t_pFirst;
        t_pFirst = t_pSecond;
        t_pSecond = t_pSecond->m_pNext;
    }
    t_pFirst->m_pNext = in_pListNode;
    return t_pFirst;
}

int main() {
    ListNode* t_iListNode = new ListNode();
    ListNode* t_iListNodeHead = t_iListNode;
    for(int i = 1; i < 10; i ++) {
        t_iListNode->m_iData = i;
        ListNode* t_iListNodeNext = new ListNode();
        t_iListNode->m_pNext = t_iListNodeNext;
        t_iListNode = t_iListNodeNext;
    }
    t_iListNode->m_iData = 10;
    ListNode* t_iTempListNode =t_iListNodeHead;
    cout << "链表翻转前：" ;
    while(t_iTempListNode != NULL) {
        cout << t_iTempListNode->m_iData << " ";
        t_iTempListNode = t_iTempListNode->m_pNext;
    }
    cout << endl;
    
    t_iTempListNode = Reverse(t_iListNodeHead);
    
    cout << "链表翻转后：" ;
    while(t_iTempListNode != NULL) {
        cout << t_iTempListNode->m_iData << " ";
        t_iTempListNode = t_iTempListNode->m_pNext;
    }
    cout << endl;
      system("pause"); 
      return 0;
     
     
}