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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;   }

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