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

//使用了三个指针，每次修改pnext指向pcurrent，然后移动三个指针#include<iostream>using namespace std;struct Node{int data;Node* next;};int a[5]={1,2,3,4,5};int main(){Node *head;head=(Node *)malloc(sizeof(Node));head->next=NULL;for(int i=0;i<5;i++){Node *tmp=(Node *)malloc(sizeof(Node));tmp->data=a[i];tmp->next=head->next;head->next=tmp;}for(Node *tmp=head->next;tmp;tmp=tmp->next)cout<<tmp->data<<" ";cout<<endl;Node* pcurrent,*pnext,*pnnext;pcurrent=(Node *)malloc(sizeof(Node));pnext=(Node *)malloc(sizeof(Node));pnnext=(Node *)malloc(sizeof(Node));pcurrent=head->next;pnext=pcurrent->next;pnnext=pnext->next;int n=0;while(pnnext!=NULL){n++;if(n==1)pcurrent->next=NULL;pnext->next=pcurrent;pcurrent=pnext;pnext=pnnext;pnnext=pnnext->next;}if(pnext->next!=NULL)pnext->next=pcurrent;head->next=pnext;for(Node *tmp=head->next;tmp;tmp=tmp->next)cout<<tmp->data<<" ";return 0;}