我是建立一个链表，然后存入数据，再逆序输出数据
#include<stdio.h>
struct Node
{ int data;
  struct Node *next;
}*Head,*p,*q;
typedef struct Node *LinkList;
void main()
{ int i,n=0;
  int idata;
  printf("输入0结束\n");
  scanf("%d",&idata);
  while(idata!=0)
  { if(n==0){p=(LinkList)malloc(sizeof(struct Node));
    p->data=idata;
    q=p;
   
    n++;}
  else {p=(LinkList)malloc(sizeof(struct Node));
    p->data=idata;
    p->next =q;
   
    n++;
  
  }
    scanf("%d",&idata);
    p=p->next ;
  }
  Head=q;
  for(i=0;i<n;i++)
  {   if(i!=n-1)
  {
   printf("%d-->",Head->data);   Head=Head->next;
  }
  else
  {
   printf("Null");  
  }

  free(q);
  q=Head;
 }
 printf("\n");
}
 

最快的辦法，是構造雙向鏈表。

#include<stdio.h>
#include<stdlib.h>       //我是这么写的
#include<string.h>
struct node 
{
    int date;
    struct node * pnext;
}; 
struct node * creat_list(struct node * h);  //创建链表
struct node * change_list(struct node * h);//调用函数进行反转 
void print_list(struct node * h);         //输出链表
int main()
{
    node * head;
    head = NULL;
    printf("请输入数据：\n");
    head = creat_list(head);
    printf("-------------------------------\n");
    print_list(head);
    printf("-------------------------------\n");
    head = change_list(head);  
    print_list(head);
} 
struct node * creat_list(struct node * h)
{
    node * p1,* p2;
     p1 = p2 = (node *)malloc(sizeof(node));
     if( p2 != NULL)
     {
         scanf("%d",&p2->date);
         p2->pnext = NULL;
     }
     while( p2->date != 0)
     {
         if(h == NULL)
             h = p2;
         else
             p1->pnext = p2;
         p1 = p2;
         p2 = (node *)malloc(sizeof(node));
         if( p2 != NULL)
         {
             scanf("%d",&p2->date);
             p2->pnext = NULL;
         }
     }
     return h;
}
void print_list(struct node * h)
{
    struct node * temp;
    temp = h;
    while(temp != NULL)
    {
        printf("%-4d\n",temp->date);
        temp = temp->pnext;
    }
}
struct node *change_list(struct node * h)
{   
    if(h == NULL||h->pnext == NULL)
          return h; 
    node * p,* q,* r;
    p = h;
    q = h->pnext;
    h->pnext = NULL;
    while(q != NULL)
    {
        r = q->pnext;
        q->pnext = p;
        p = q;
        q = r;
    }
    h = p;
    return h; 
}

虽然建议模块化编程, 但楼主的功能,使用头插法就可以了.

对哈！哈哈

你試試

#include<stdio.h>
#include<stdlib.h>                     //是这样的吧？
struct Node
{ 
    int data;
    struct Node *next;
};
void print_list(struct Node * h);     //输出链表函数
void main()
{
    struct Node * head,* p1,* p2;
    head = (struct Node *)malloc(sizeof(struct Node));
    scanf("%d",&head -> data);
    head -> next = NULL;
    while(head -> data != 0)
    {
        p1 = (struct Node *)malloc(sizeof(struct Node));
        p1 -> next = head;
        scanf("%d",&p1 -> data);
        head = p1;
    }
    print_list(head);
}
void print_list(struct Node * h)
{
    struct Node * temp;
    temp = h -> next;
    while(temp != NULL)
    {
        printf("%d\n",temp -> data);
        temp = temp -> next;
    }
}