程序代码：

#include<stdio.h>
#include<stdlib.h>
#define N 20
/* User Code Begin(Limit: lines<=5, lineLen<=60, 考生可在本行后添加代码、仅5行、行长<=60字符，功能是节点类型Link定义) */
struct Link {
int data;
struct Link *next;
};


/* User Code End(考生添加代码结束) */
void dispLink(struct Link *Head);  /* 根据给定的链首Head，显示整个链表中的数据，同时检查两次使用的链表是否刚好反序 */
/* User Code Begin(Limit: lines<=2, lineLen<=60, 考生可在本行后添加代码、仅2行、行长<=60字符，功能是用户自定义函数的原型声明) */
struct Link * reverseLink(struct Link * head );
struct Link * creatLink();
/* User Code End(考生添加代码结束) */
int main(void)
{
    struct Link *Head;
    Head = creatLink();  /* 创建新链表 */
    printf("\nbefore:");
    dispLink(Head);
    Head = reverseLink(Head);  /* 反转链表 */
    printf("\nafter:");
    dispLink(Head);
    return 0;
}
void dispLink(struct Link *Head)
{
    static struct Link *oLink[2][N];
    static int callNumber = -1;
    int i = 0;
    callNumber++;
    if (callNumber > 1)
    {
        printf("Error, call invalid!\n");
        return ;
    }
   
    while (Head != NULL)
    {
        oLink[callNumber][i] = Head;
        i++;
        printf("%d ", Head->data);
        Head = Head->next;
    }
    printf("\n");
    if (1 == callNumber)
    {
        for (i=0; i<N; i++)
        {
            if (oLink[0][i] != oLink[1][N-1-i])
            {
                printf("Error, Link not reverse!\n");
                return ;
            }
        }
    }
}
/* User Code Begin(考生在此后根据设计需要完成程序的其它部分，行数不限) */
struct Link * creatLink()
{
struct Link * head,*p1,*p2;
int i;
head=NULL;
p1=p2=(struct Link *)malloc(sizeof(struct Link));
  for(i=0;i<20;i++)
  {
      p1->data=i;
      if(i=0)head=p1;
      else p2->next=p1;
      p2=p1;
      p1=(struct Link *)malloc(sizeof(struct Link));
  }
  p2->next=NULL;
  return head;
}
struct Link * reverseLink(struct Link * head )
{
struct Link *p1,*p2,*t;
if(head!=NULL)p1=head;
for(;p1!=NULL;)
{
p1=p1->next;
}
t=p1->next;
p1->next=head;
head=t;
return head;
}