Write a C Program to check linked list is Palindrome or not

By | 08.04.2017

Check linked list is Palindrome or not


Write a C Program to check linked list is Palindrome or not. Here’s simple Program to check linked list is Palindrome or not in C Programming Language.


What is Linked List ?


Linked list is a linear data structure that contains sequence of elements such that each element links to its next element in the sequence. Each link contains a connection to another link.

Following are the important terms to understand the concept of Linked List.

  • Link − Each link of a linked list can store a data called an element.
  • Next − Each link of a linked list contains a link to the next link called Next.

Each element in a linked list is called as “Node”. Each node consists of its own data and the address of the next node and forms a chain. Linked Lists are used to create trees and graphs.


Below is the source code for C Program to check linked list is Palindrome or not which is successfully compiled and run on Windows System to produce desired output as shown below :


SOURCE CODE : :


/* C Program to check linked list is Palindrome or not */

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

struct node
{
        int info;
        struct node *link;
};

struct node *create_list(struct node *start);
void display(struct node *start);
struct node *addatbeg(struct node *start,int data);
struct node *addatend(struct node *start,int data);
int isIdentical(struct node *p1, struct node *p2);
int isPalindrome(struct node *start);
struct node *reverse(struct node *start);

int main()
{
        struct node *start=NULL;
        start=create_list(start);
        display(start);
        if( isPalindrome(start) )
                printf("\nList is a Palindrome\n");
        else
                printf("\nList is not a Palindrome\n");

        return 0;

}/*End of main()*/

int isPalindrome(struct node *start)
{
        struct node *slowP, *fastP, *prev_slowP, *start_second;
        int result;

        if(start==NULL || start->link==NULL) /*List empty or only one element*/
           return 1;
        prev_slowP = slowP = fastP = start;
        while(fastP->link!=NULL && fastP->link->link!=NULL)
        {
                prev_slowP = slowP;
                slowP=slowP->link;
                fastP=fastP->link->link;
        }
        start_second = slowP->link;
        if(fastP->link == NULL)  /*odd number of elements in the list*/
    {
                start_second = reverse(start_second);  /*reverse the second half*/
                prev_slowP->link = NULL;
                result = isIdentical(start, start_second);/*compare the first half and second half*/
                start_second = reverse(start_second);  /*reverse second half to get the original order*/
                prev_slowP->link = slowP; /*Join first half with the middle element*/
                slowP->link = start_second;/*Join middle element with the second half*/
    }
        else /*even number of elements in the list*/
        {
                start_second = reverse(start_second);  /*reverse the second half*/
                slowP->link = NULL; /*node pointed by slowP is the last node of first half*/
                result = isIdentical(start, start_second);  /*compare the first half and second half*/
                start_second = reverse(start_second);  /*reverse second half to get the original order*/
                slowP->link = start_second; /*Join first half and second half*/
        }
        return result;
}/*End of isPalindrome()*/

struct node *reverse(struct node *start)
{
        struct node *prev, *ptr, *next;
        prev=NULL;
        ptr=start;
        while(ptr!=NULL)
        {
                next=ptr->link;
                ptr->link=prev;
                prev=ptr;
                ptr=next;
        }
        start=prev;
        return start;
}/*End of reverse()*/

int isIdentical(struct node *p1, struct node *p2)
{
        while(1)
        {
                if((p1==NULL)&&(p2==NULL))
                        return 1;
                if((p1==NULL)||(p2==NULL))
                        return 0;
                if(p1->info!=p2->info)
                        return 0;
                p1 = p1->link;
                p2 = p2->link ;
        }
}/*End of Identical()*/

struct node *create_list(struct node *start)
{
        int i,n,data;
        printf("Enter the number of nodes : ");
        scanf("%d",&n);
        start=NULL;
        if(n==0)
                return start;

        printf("\nEnter the element to be inserted : ");
        scanf("%d",&data);
        start=addatbeg(start,data);

        for(i=2;i<=n;i++)
        {
                printf("\nEnter the element to be inserted : ");
                scanf("%d",&data);
                start=addatend(start,data);
        }
        return start;
}/*End of create_list()*/

void display(struct node *start)
{
        struct node *p;
        if(start==NULL)
        {
                printf("\nList is empty\n");
                return;
        }
        p=start;
        printf("\nList is :\n");
        while(p!=NULL)
        {
                printf("%d ",p->info);
                p=p->link;
        }
        printf("\n\n");
}/*End of display() */

struct node *addatbeg(struct node *start,int data)
{
        struct node *tmp;
        tmp=(struct node *)malloc(sizeof(struct node));
        tmp->info=data;
        tmp->link=start;
        start=tmp;
        return start;
}/*End of addatbeg()*/

struct node *addatend(struct node *start,int data)
{
        struct node *p,*tmp;
        tmp=(struct node *)malloc(sizeof(struct node));
        tmp->info=data;
        p=start;
        while(p->link!=NULL)
                p=p->link;
        p->link=tmp;
        tmp->link=NULL;
        return start;
}/*End of addatend()*/

OUTPUT : :


/* C Program to check linked list is Palindrome or not */

************** OUTPUT ****************

Enter the number of nodes : 6

Enter the element to be inserted : 2

Enter the element to be inserted : 4

Enter the element to be inserted : 5

Enter the element to be inserted : 5

Enter the element to be inserted : 4

Enter the element to be inserted : 2

List is :
2 4 5 5 4 2


List is a Palindrome

If you found any error or any queries related to the above program or any questions or reviews , you wanna to ask from us ,you may Contact Us through our contact Page or you can also comment below in the comment section.We will try our best to reach up to you in short interval.


Thanks for reading the post….

4.5 2 votes
Article Rating
Subscribe
Notify of
guest

0 Comments
Inline Feedbacks
View all comments