实验11-2-9 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：
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| struct ListNode {
int data;
struct ListNode *next;
};
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函数接口定义：
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| struct ListNode *reverse( struct ListNode *head );
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其中head是用户传入的链表的头指针；函数reverse将链表head逆置，并返回结果链表的头指针。
裁判测试程序样例：
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| #include <stdio.h>
#include <stdlib.h>
struct ListNode {
int data;
struct ListNode *next;
};
struct ListNode *createlist();
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;
}
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输入样例：
输出样例：
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| #include <stdio.h>
#include <stdlib.h>
struct ListNode {
int data;
struct ListNode *next;
};
struct ListNode *createlist();
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()
{
struct ListNode *p,*h,*q;
p=q=(struct ListNode*)malloc(sizeof(struct ListNode));
h=(struct ListNode*)malloc(sizeof(struct ListNode));
int data;
h=p;
scanf("%d",&data);
while(data!=-1)
{
p->data=data;
q->next=p;
q=p;
p=(struct ListNode*)malloc(sizeof(struct ListNode));
scanf("%d",&data);
}
q->next=NULL;
free(p);
return h;
}
struct ListNode *reverse( struct ListNode *head )
{
struct ListNode *p,*q=NULL,*r=NULL;
if(head==NULL)
{
return NULL;
}
if(head->next==NULL)
{
return head;
}
p=head;
while(p!=NULL)
{
q=p->next;
p->next=r;
r=p;
p=q;
}
return r;
}
TmpCell = head;
head = head->next;
free(TmpCell);
return head;
}
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