Check if a linked list of strings forms a palindrome

Given a linked list handling string data, check to see whether data is palindrome or not? 

For example, 

Input  : a -> bc -> d -> dcb -> a -> NULL
Output : True
String "abcddcba" is palindrome.

Output : a -> bc -> d -> ba -> NULL
Output : False
String "abcdba" is not palindrome. 

The idea is very simple. Construct a string out of given linked list and check if the constructed string is palindrome or not. 

C++

// Program to check if a given linked list of strings
// form a palindrome
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
struct Node
{
    string data;
    Node* next;
};

// A utility function to check if str is palindrome
// or not
bool isPalindromeUtil(string str)
{
    int length = str.length();

    // Match characters from beginning and
    // end.
    for (int i=0; i<length/2; i++)
        if (str[i] != str[length-i-1])
            return false;

    return true;
}

// Returns true if string formed by linked
// list is palindrome
bool isPalindrome(Node *node)
{
    // Append all nodes to form a string
    string str = "";
    while (node != NULL)
    {
        str.append(node->data);
        node = node->next;
    }

    // Check if the formed string is palindrome
    return isPalindromeUtil(str);
}

// A utility function to print a given linked list
void printList(Node *node)
{
    while (node != NULL)
    {
        cout << node->data << " -> ";
        node = node->next;
    }
    printf("NULL\n");
}

/* Function to create a new node with given data */
Node *newNode(const char *str)
{
    Node *new_node = new Node;
    new_node->data = str;
    new_node->next = NULL;
    return new_node;
}

/* Driver program to test above function*/
int main()
{
    Node *head = newNode("a");
    head->next = newNode("bc");
    head->next->next = newNode("d");
    head->next->next->next = newNode("dcb");
    head->next->next->next->next = newNode("a");

    isPalindrome(head)? printf("true\n"):
                        printf("false\n");

    return 0;
}

Java

// Java Program to check if a given linked list of strings
// form a palindrome

import java.util.Scanner;

// Linked List node
class Node
{
    String data;
    Node next;

    Node(String d)
    {
        data = d;
        next = null;
    }
}

class LinkedList_Palindrome
{
    Node head;

    // A utility function to check if str is palindrome
    // or not
    boolean isPalidromeUtil(String str)
    {
        int length = str.length();

        // Match characters from beginning and
        // end.
        for (int i=0; i<length/2; i++)
            if (str.charAt(i) != str.charAt(length-i-1))
                return false;

        return true;
    }

    // Returns true if string formed by linked
    // list is palindrome
    boolean isPalindrome()
    {
        Node node = head;

        // Append all nodes to form a string
        String str = "";
        while (node != null)
        {
            str = str.concat(node.data);
            node = node.next;
        }

        // Check if the formed string is palindrome
        return isPalidromeUtil(str);
    }

    /* Driver program to test above function*/
    public static void main(String[] args)
    {
        LinkedList_Palindrome list = new LinkedList_Palindrome();
        list.head = new Node("a");
        list.head.next = new Node("bc");
        list.head.next.next = new Node("d");
        list.head.next.next.next = new Node("dcb");
        list.head.next.next.next.next = new Node("a");

        System.out.println(list.isPalindrome());

    }
}
// This code has been contributed by Amit Khandelwal

Python3

# Python program to check if given linked list of
# strings form a palindrome

# Node class
class Node:

    # Constructor to initialize the node object
    def __init__(self, data):
        self.data = data
        self.next = None

class LinkedList:

    # Function to initialize head
    def __init__(self):
        self.head = None

    # A utility function to check if str is palindrome
    # or not
    def isPalindromeUtil(self, string):
        return (string == string[::-1])

    # Returns true if string formed by linked list is
    # palindrome
    def isPalindrome(self):
        node = self.head

        # Append all nodes to form a string
        temp = []
        while (node is not None):
            temp.append(node.data)
            node = node.next
        string = "".join(temp)
        return self.isPalindromeUtil(string)

    # Utility function to print the linked LinkedList
    def printList(self):
        temp = self.head
        while (temp):
            print(temp.data,end=" ")
            temp = temp.next


# Driver program to test above function
llist = LinkedList()
llist.head = Node('a')
llist.head.next = Node('bc')
llist.head.next.next = Node("d")
llist.head.next.next.next = Node("dcb")
llist.head.next.next.next.next = Node("a")
print ("true" if llist.isPalindrome() else "false")

# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

C#

// C# Program to check if a given linked list
// of strings form a palindrome
using System;

// Linked List node
class Node
{
    public String data;
    public Node next;

    public Node(String d)
    {
        data = d;
        next = null;
    }
}

public class LinkedList_Palindrome
{
    Node head;

    // A utility function to check if
    // str is palindrome or not
    bool isPalidromeUtil(String str)
    {
        int length = str.Length;

        // Match characters from beginning and
        // end.
        for (int i = 0; i < length / 2; i++)
            if (str[i] != str[length - i - 1])
                return false;

        return true;
    }

    // Returns true if string formed by linked
    // list is palindrome
    bool isPalindrome()
    {
        Node node = head;

        // Append all nodes to form a string
        String str = "";
        while (node != null)
        {
            str = str+(node.data);
            node = node.next;
        }

        // Check if the formed string is palindrome
        return isPalidromeUtil(str);
    }

    /* Driver code*/
    public static void Main(String[] args)
    {
        LinkedList_Palindrome list = new LinkedList_Palindrome();
        list.head = new Node("a");
        list.head.next = new Node("bc");
        list.head.next.next = new Node("d");
        list.head.next.next.next = new Node("dcb");
        list.head.next.next.next.next = new Node("a");

        Console.WriteLine(list.isPalindrome());
    }
}

// This code has been contributed
// by PrinciRaj1992

Output

true

Time Complexity: O(n), where n is the number of nodes in the given linked list.
Auxiliary Space: O(m), where m is the length of the string formed by the linked list.

This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.