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 \n",p->index);
    }
        return 0;
}