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*/

}

#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 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; 

}