单链表就地反转

　　实现一个函数:void reverse(struct list_node *head)在尽量不借助辅助变量的情况下，实现任意长度单链表（不考虑内存限制）的反转（or 逆序）。

struct list_node{
	int val;
	struct list_node *next;
};

struct list{
	struct list_node *head;	
	struct list_node *tail;
};

void reverse(struct list_node *head)
{

}
int main()
{
    struct list list = {NULL, NULL};
    struct list_node *p = NULL;
    /*init list*/
    /*打印反转前各节点的值*/
    reverse(list.head);
    p = list.head;
    list.head = list.tail;
    list.tail = p;
    /*打印反转后各节点的值*/
}

　　代码实现：

#include <stdio.h>

struct list_node{
	int index;
	struct list_node *next;
};
struct list
{
	struct list_node *head;
	struct list_node *tail;
};
void reverse(struct list_node *head)
{
	if(NULL == head|| NULL == head->next )
	    return;
	reverse(head->next);
	head->next->next = head;
	head->next = NULL;
}
int main()
{
	int i = 0;
	struct list list = {NULL, NULL};
	struct list_node node[10] = {0};
	struct list_node *p;
	for(i = 0; i < 9; i++)
	{
		node[i].index = i;
		node[i].next = &node[i + 1];
	}
	node[9].index = 9;
	list.head = node;
	list.tail = node + 9;
	
	reverse(list.head);
	for(p = list.tail; p; p=p->next)
	{
		printf("%d 
",p->index);
	}
        return 0;  
}