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:
struct ListNode {
int data;
struct ListNode *next;
};
函数接口定义:
struct ListNode *reverse( struct ListNode *head );
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:
#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;
}
输入样例:
1 2 3 4 5 6 -1
输出样例:
6 5 4 3 2 1
struct ListNode *reverse( struct ListNode *head )
{
struct ListNode *p = NULL,*p1 = NULL;
while(head != NULL)
{
p=(struct ListNode*)malloc(sizeof(struct ListNode));
p -> data = head -> data;
p -> next = p1;
p1 = p;
head = head -> next;
}
return p;
}