利用栈来实现单链表的逆序

#include <stdio.h>
# include<stdlib.h>
#define stacksize 100
typedef int datatype;
typedef struct
{
   datatype data[stacksize];
        int  top;
}seqstack;
typedef struct node{
   datatype  data;
   struct node *next;
}listnode;
typedef listnode *linklist;
linklist head;
linklist p;
int count;
linklist creatlist(int n)
{
  linklist head;
  listnode *p1,*p2;
  int i;
  head=(linklist)malloc(sizeof(listnode));
  head->next=NULL;
  p2=head;
  printf("Please input the records of the chain!\n");
  for(i=0;i<n;i++)
  {
     p1=(linklist)malloc(sizeof(listnode));
     scanf("%d",&p1->data);
     p1->next=p2->next;
     p2->next=p1;
     p2=p1;
  }
   return (head);
}
void print(linklist head,int n)
{
  if(head==NULL)
           printf("list null!\n");
  printf("Now these %d records are:\n",n);
  p=head->next;
  printf("%4d",p->data);
  count=1;
  while(p->next!=NULL)
  {
      count++;
      p=p->next;
      printf("%4d",p->data);
      if(!(count%10))
             printf("\n");
  }
}
datatype push(seqstack *s,int x)  
 {
  if(s->top==stacksize-1)
       printf("The stack is overflow!\n");
   else
             s->data[++s->top]=x;  
 }
datatype pop(seqstack *s) 
 {
   if(s->top==-1)
    printf("the stack is empty!\n");
   else
           return (s->data[s->top--]);
 }
datatype deleted(linklist head)
{
    datatype temp;
    linklist p;
    p=head->next;
    temp=(p->data);
    head->next=p->next;
    free(p);
    return (temp);
}
void invertedlist(linklist head,int n)
{   
    seqstack s;
    int i,j,temp;
    s.top=-1;
    for(i=0;i<n;i++)
    {
       temp=deleted(head);
       push(&s,temp); 
    }
    for(i=0;i<n;i++)
    {
       temp=pop(&s);
       printf("%5d",temp);
       if(!((i+1)%10))
                printf("\n");
    }
    printf("\n");
}
main()
{
    linklist head;
    int n;
    printf("Please input the value of n:\n");
    scanf("%d",&n);
    head=creatlist(n);
    print(head,n);
    printf("\n");
    printf("After inverting,the records of the chain are:\n");
    invertedlist(head,n);
    system("pause");
    return 0;
} 

运行结果：   
Please input the value of n:
10
Please input the records of the chain!
-5 8 6 -1 0 1 2 -5 6 7
Now these 10 records are:
  -5   8   6  -1   0   1   2  -5   6   7

After inverting,the records of the chain are:
   7    6   -5    2    1    0   -1    6    8   -5

请按任意键继续. . .