Check if a doubly linked list of characters is palindrome or not

Given a doubly-linked list of characters, write a function that returns true if the given doubly linked list is a palindrome, else false. 

1. Create a doubly linked list where each node contains only one character of a string.
2. Initialize two pointers left at the start of the list and right at the end of the list.
3. Check if the data on the left node is equal to the right node. If it is equal, then increment left and decrement right till the middle of the list. If, at any stage, it is not equal, then return false.

Implementation:

C++

// C++ program to check if doubly
// linked list is palindrome or not
#include<bits/stdc++.h>
using namespace std;

// Structure of node
struct Node
{
    char data;
    struct Node *next;
    struct Node *prev;
};

/* Given a reference (pointer to pointer) to
   the head of a list and an int, inserts a
   new node on the front of the list. */
void push(struct Node** head_ref, char new_data)
{
    struct Node* new_node = new Node;
    new_node->data  = new_data;
    new_node->next = (*head_ref);
    new_node->prev = NULL;
    if ((*head_ref) !=  NULL)
      (*head_ref)->prev = new_node ;
    (*head_ref)    = new_node;
}

// Function to check if list is palindrome or not
bool isPalindrome(struct Node *left)
{
    if (left == NULL)
       return true;

    // Find rightmost node
    struct Node *right = left;
    while (right->next != NULL)
        right = right->next;

    while (left != right)
    {
        if (left->data != right->data)
            return false;

        left = left->next;
        right = right->prev;
    }

    return true;
}

// Driver program
int main()
{
    struct Node* head = NULL;
    push(&head, 'l');
    push(&head, 'e');
    push(&head, 'v');
    push(&head, 'e');
    push(&head, 'l');

    if (isPalindrome(head))
        printf("It is Palindrome");
    else
        printf("Not Palindrome");

    return 0;
}

Java

// Java program to check if doubly
// linked list is palindrome or not
class GFG
{

// Structure of node
static class Node
{
    char data;
    Node next;
    Node prev;
};

/* Given a reference (pointer to pointer) to
the head of a list and an int, inserts a
new node on the front of the list. */
static Node push(Node head_ref, char new_data)
{
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = head_ref;
    new_node.prev = null;
    if (head_ref != null)
    head_ref.prev = new_node ;
    head_ref = new_node;
    return head_ref;
}

// Function to check if list is palindrome or not
static boolean isPalindrome(Node left)
{
    if (left == null)
    return true;

    // Find rightmost node
    Node right = left;
    while (right.next != null)
        right = right.next;

    while (left != right)
    {
        if (left.data != right.data)
            return false;

        left = left.next;
        right = right.prev;
    }

    return true;
}

// Driver program
public static void main(String[] args)
{
    Node head = null;
    head = push(head, 'l');
    head = push(head, 'e');
    head = push(head, 'v');
    head = push(head, 'e');
    head = push(head, 'l');

    if (isPalindrome(head))
        System.out.printf("It is Palindrome");
    else
        System.out.printf("Not Palindrome");
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 program to check if doubly
# linked list is a palindrome or not

class Node:

    def __init__(self, data, next, prev):
        self.data = data
        self.next = next
        self.prev = prev

# Given a reference (pointer to pointer) to
# the head of a list and an int, inserts
# a new node on the front of the list.
def push(head_ref, new_data):

    new_node = Node(new_data, head_ref, None)

    if head_ref != None:
        head_ref.prev = new_node
        head_ref = new_node

    return head_ref

# Function to check if list is palindrome or not
def isPalindrome(left):

    if left == None:
        return True

    # Find rightmost node
    right = left
    while right.next != None:
        right = right.next

    while left != right:

        if left.data != right.data:
            return False

        left = left.next
        right = right.prev

    return True

# Driver program
if __name__ == "__main__":

    head = None
    head = push(head, 'l')
    head = push(head, 'e')
    head = push(head, 'v')
    head = push(head, 'e')
    head = push(head, 'l')

    if isPalindrome(head):
        print("It is Palindrome")
    else:
        print("Not Palindrome")

# This code is contributed by Rituraj Jain

C#

// C# program to check if doubly
// linked list is palindrome or not
using System;

class GFG
{

// Structure of node
public class Node
{
    public char data;
    public Node next;
    public Node prev;
};

/* Given a reference (pointer to pointer) to
the head of a list and an int, inserts a
new node on the front of the list. */
static Node push(Node head_ref, char new_data)
{
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = head_ref;
    new_node.prev = null;
    if (head_ref != null)
    head_ref.prev = new_node ;
    head_ref = new_node;
    return head_ref;
}

// Function to check if list is palindrome or not
static bool isPalindrome(Node left)
{
    if (left == null)
    return true;

    // Find rightmost node
    Node right = left;
    while (right.next != null)
        right = right.next;

    while (left != right)
    {
        if (left.data != right.data)
            return false;

        left = left.next;
        right = right.prev;
    }

    return true;
}

// Driver program
public static void Main(String[] args)
{
    Node head = null;
    head = push(head, 'l');
    head = push(head, 'e');
    head = push(head, 'v');
    head = push(head, 'e');
    head = push(head, 'l');

    if (isPalindrome(head))
        Console.Write("It is Palindrome");
    else
        Console.Write("Not Palindrome");
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>
    // javascript program to check if doubly
    // linked list is palindrome or not    // Structure of node
    class Node {
        constructor() {
            this.data = 0;
            this.prev = null;
            this.next = null;
        }
    }

    /*
     * Given a reference (pointer to pointer) to the head of a list and an int,
     * inserts a new node on the front of the list.
     */
    function push(head_ref,  new_data) {
        var new_node = new Node();
        new_node.data = new_data;
        new_node.next = head_ref;
        new_node.prev = null;
        if (head_ref != null)
            head_ref.prev = new_node;
        head_ref = new_node;
        return head_ref;
    }

    // Function to check if list is palindrome or not
    function isPalindrome(left) {
        if (left == null)
            return true;

        // Find rightmost node
        var right = left;
        while (right.next != null)
            right = right.next;

        while (left != right) {
            if (left.data != right.data)
                return false;

            left = left.next;
            right = right.prev;
        }

        return true;
    }

    // Driver program

    var head = null;
    head = push(head, 'l');
    head = push(head, 'e');
    head = push(head, 'v');
    head = push(head, 'e');
    head = push(head, 'l');

    if (isPalindrome(head))
        document.write("It is Palindrome");
    else
        document.write("Not Palindrome");

// This code contributed by umadevi9616
</script>

Output

It is Palindrome

Time complexity: O(n) 
Auxiliary Space : O(1)

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