链表倒序

 现在不说废话了.第一个取下的结点next指针指向NULL,然后依次取下的结点的next指针指向前一个取下的结点,而前一个取下的结点prev指针指向当前取下的结点就行了,一次遍历链表就搞定了.

 

#include <stdio.h>
#include <stdlib.h>

typedef struct node{
        int num;
        struct node* prev;
        struct node* next;
        }NODE;
        
NODE* create_list(int n)
{
      int i = 0;
      NODE* p = NULL;
      NODE* q = NULL;
      NODE* head = (NODE*)malloc(sizeof(NODE));
      
      p = head;
      
      for(i=0;i<n;i++)
       {
          q = (NODE*)malloc(sizeof(NODE));
          q->num = i;
          p->next = q;
          q->next = NULL;
          p = q;
       }
       
       return head;
}

NODE* revert(NODE* head)
{
      NODE* p1 = head->next;
      NODE* p2 = NULL;
      NODE* p3 = NULL;
      
      //p3->next =NULL;
      while(p1!=NULL)
       {
         printf("now %d\n",p1->num);
         p2 = p1;
         p1 = p1->next;
         p2->next = p3;
         p3 = p2;
         }   
         head->next = p3;
}
      
int my_print(NODE* head)
{
    NODE* p = head->next;
    while(p!=NULL)
     {
          printf("%d\t",p->num);
          p = p->next;
     }
     printf("\n");
}
          
int main(int argc, char *argv[])
{
  NODE* head = create_list(5);
  my_print(head);
  revert(head);
  my_print(head);
  
  system("PAUSE");    
  return 0;
}
代码是在dev c++上写的,gcc估计system那句要改^_^