反向显示双向链表

实现

该算法的实现如下 −

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

struct node {
   int data;
   struct node *prev;
   struct node *next;
};

struct node *head = NULL;
struct node *last = NULL;

struct node *current = NULL;

//display the list
void printList() {
   struct node *ptr = head;

   printf("
[head] <=>");
   //从头开始
   while(ptr != NULL) {        
      printf(" %d <=>",ptr->data);
      ptr = ptr->next;
   }

   printf(" [last]
");
}

//display the list
void print_backward() {
   struct node *ptr = last;

   printf("
[head] <=>");
   //从头开始
   while(ptr != NULL) {        
      printf(" %d <=>",ptr->data);
      ptr = ptr->prev;
   }

   printf(" [last]
");
}

//创建链接列表
void insert(int data) {
   // 为新节点分配内存;
   struct node *link = (struct node*) malloc(sizeof(struct node));

   link->data = data;
   link->prev = NULL;
   link->next = NULL;

   // 如果 head 为空，则创建新列表
   if(head==NULL) {
      head = link;
      return;
   }

   current = head;
   
   // 移至列表末尾
   while(current->next!=NULL)
      current = current->next;

   // 在列表末尾插入链接
   current->next = link;
   last = link;
   link->prev = current;
}

int main() {
   insert(10);
   insert(20);
   insert(30);
   insert(1);
   insert(40);
   insert(56); 

   printList();
   print_backward();
   
   return 0;
}

输出

程序的输出应为 −

[head] <=> 10 <=> 20 <=> 30 <=> 1 <=> 40 <=> 56 <=> [last] 

[head] <=> 56 <=> 40 <=> 1 <=> 30 <=> 20 <=> 10 <=> [last]

linked_list_programs_in_c.html