Display Doubly Linked List In Reverse

Implementation

Implementation of this algorithm is given below −

#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("\n[head] <=>");
   //start from the beginning
   while(ptr != NULL) {        
      printf(" %d <=>",ptr->data);
      ptr = ptr->next;
   }

   printf(" [last]\n");
}

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

   printf("\n[head] <=>");
   //start from the beginning
   while(ptr != NULL) {        
      printf(" %d <=>",ptr->data);
      ptr = ptr->prev;
   }

   printf(" [last]\n");
}

//Create Linked List
void insert(int data) {
   // Allocate memory for new node;
   struct node *link = (struct node*) malloc(sizeof(struct node));

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

   // If head is empty, create new list
   if(head==NULL) {
      head = link;
      return;
   }

   current = head;
   
   // move to the end of the list
   while(current->next!=NULL)
      current = current->next;

   // Insert link at the end of the list
   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;
}

Output

Output of the program should be −

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

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

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