6-3 链表逆置 (20 分) 本题要求实现一个函数，将给定单向链表逆置，即表头置为表尾，表尾置为表头。链表结点定义如下：

```c
#include <stdio.h>
#include <stdlib.h>

struct ListNode {
    int data;
    struct ListNode* next;
};

struct ListNode* createlist(); /*裁判实现，细节不表*/
struct ListNode* reverse(struct ListNode* head);
void printlist(struct ListNode* head)
{
    struct ListNode* p = head;
    while (p) {
        printf("%d ", p->data);
        p = p->next;
    }
    printf("\n");
}

int main()
{
    struct ListNode* head;

    head = createlist();
    head = reverse(head);
    printlist(head);

    return 0;
}


struct ListNode* createlist(void) /*裁判实现，细节不表*/
{
    struct ListNode* head = NULL, * tail = NULL, * p = NULL;
    int data;
    while (1) {
        scanf("%d", &data);
        if (data == -1)
            return head;
        p = (struct ListNode*)malloc(sizeof(struct ListNode));
        p->data = data;
        p->next = NULL;
        if (head) {
            tail->next = p;
            tail = p;
        }
        else {
            head = p;
            tail = p;
        }
    }
}
/* 你的代码将被嵌在这里 */
struct ListNode* reverse(struct ListNode* head)
{
    struct ListNode* p = head;
    struct ListNode* newhead = NULL;
   while (p)
    {
        
       struct ListNode* abc = (struct ListNode*)malloc(sizeof(struct ListNode));
       abc->data = p->data; abc->next=NULL;
       if (newhead == NULL)
           newhead = abc;
       else
       {
           abc->next = newhead;
           newhead = abc;
       }
        p = p->next;
      }
   return newhead;
}