Function to check if a singly linked list is palindrome

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

Examples:

Input: R->A->D->A->R->NULL
Output: Yes

Input: C->O->D->E->NULL
Output: No

Check if a Singly Linked List is Palindrome using Stack:

The idea is to use a stack and push all the nodes into the stack, then again iterate over the linked list to validate if the linked list is palindrome or not.

Below image is a dry run of the above approach: 

Follow the steps below to solve the problem:

• A simple solution is to use a stack of list nodes. This mainly involves three steps.
• Traverse the given list from head to tail and push every visited node to stack.
• Traverse the list again. For every visited node, pop a node from the stack and compare data of popped node with the currently visited node.
• If all nodes matched, then return true, else false.

Below is the implementation of the above approach : 

C++

#include <bits/stdc++.h>
using namespace std;

class Node {
public:
    int data;
    Node(int d) { data = d; }
    Node* ptr;
};

// Function to check if the linked list
// is palindrome or not
bool isPalin(Node* head)
{

    // Temp pointer
    Node* slow = head;

    // Declare a stack
    stack<int> s;

    // Push all elements of the list
    // to the stack
    while (slow != NULL) {
        s.push(slow->data);

        // Move ahead
        slow = slow->ptr;
    }

    // Iterate in the list again and
    // check by popping from the stack
    while (head != NULL) {

        // Get the top most element
        int i = s.top();

        // Pop the element
        s.pop();

        // Check if data is not
        // same as popped element
        if (head->data != i) {
            return false;
        }

        // Move ahead
        head = head->ptr;
    }

    return true;
}

// Driver Code
int main()
{

    // Addition of linked list
    Node one = Node(1);
    Node two = Node(2);
    Node three = Node(3);
    Node four = Node(2);
    Node five = Node(1);

    // Initialize the next pointer
    // of every current pointer
    five.ptr = NULL;
    one.ptr = &two;
    two.ptr = &three;
    three.ptr = &four;
    four.ptr = &five;
    Node* temp = &one;

    // Call function to check palindrome or not
    int result = isPalin(&one);

    if (result == 1)
        cout << "isPalindrome is true\n";
    else
        cout << "isPalindrome is false\n";

    return 0;
}

// This code has been contributed by Striver

Java

/* Java program to check if linked list is palindrome
 * recursively */
import java.util.*;

class linkedList {
    public static void main(String args[])
    {
        Node one = new Node(1);
        Node two = new Node(2);
        Node three = new Node(3);
        Node four = new Node(4);
        Node five = new Node(3);
        Node six = new Node(2);
        Node seven = new Node(1);
        one.ptr = two;
        two.ptr = three;
        three.ptr = four;
        four.ptr = five;
        five.ptr = six;
        six.ptr = seven;
        boolean condition = isPalindrome(one);
        System.out.println("isPalidrome :" + condition);
    }
    static boolean isPalindrome(Node head)
    {

        Node slow = head;
        boolean ispalin = true;
        Stack<Integer> stack = new Stack<Integer>();

        while (slow != null) {
            stack.push(slow.data);
            slow = slow.ptr;
        }

        while (head != null) {

            int i = stack.pop();
            if (head.data == i) {
                ispalin = true;
            }
            else {
                ispalin = false;
                break;
            }
            head = head.ptr;
        }
        return ispalin;
    }
}

class Node {
    int data;
    Node ptr;
    Node(int d)
    {
        ptr = null;
        data = d;
    }
}

Python3

# Python3 program to check if linked
# list is palindrome using stack


class Node:
    def __init__(self, data):

        self.data = data
        self.ptr = None

# Function to check if the linked list
# is palindrome or not


def ispalindrome(head):

    # Temp pointer
    slow = head

    # Declare a stack
    stack = []

    ispalin = True

    # Push all elements of the list
    # to the stack
    while slow != None:
        stack.append(slow.data)

        # Move ahead
        slow = slow.ptr

    # Iterate in the list again and
    # check by popping from the stack
    while head != None:

        # Get the top most element
        i = stack.pop()

        # Check if data is not
        # same as popped element
        if head.data == i:
            ispalin = True
        else:
            ispalin = False
            break

        # Move ahead
        head = head.ptr

    return ispalin

# Driver Code


# Addition of linked list
one = Node(1)
two = Node(2)
three = Node(3)
four = Node(4)
five = Node(3)
six = Node(2)
seven = Node(1)

# Initialize the next pointer
# of every current pointer
one.ptr = two
two.ptr = three
three.ptr = four
four.ptr = five
five.ptr = six
six.ptr = seven
seven.ptr = None

# Call function to check palindrome or not
result = ispalindrome(one)

print("isPalindrome:", result)

# This code is contributed by Nishtha Goel

C#

// C# program to check if linked list
// is palindrome recursively
using System;
using System.Collections.Generic;

class linkedList {

    // Driver code
    public static void Main(String[] args)
    {
        Node one = new Node(1);
        Node two = new Node(2);
        Node three = new Node(3);
        Node four = new Node(4);
        Node five = new Node(3);
        Node six = new Node(2);
        Node seven = new Node(1);

        one.ptr = two;
        two.ptr = three;
        three.ptr = four;
        four.ptr = five;
        five.ptr = six;
        six.ptr = seven;

        bool condition = isPalindrome(one);
        Console.WriteLine("isPalidrome :" + condition);
    }

    static bool isPalindrome(Node head)
    {
        Node slow = head;
        bool ispalin = true;
        Stack<int> stack = new Stack<int>();

        while (slow != null) {
            stack.Push(slow.data);
            slow = slow.ptr;
        }

        while (head != null) {
            int i = stack.Pop();
            if (head.data == i) {
                ispalin = true;
            }
            else {
                ispalin = false;
                break;
            }
            head = head.ptr;
        }
        return ispalin;
    }
}

class Node {
    public int data;
    public Node ptr;
    public Node(int d)
    {
        ptr = null;
        data = d;
    }
}

// This code is contributed by amal kumar choubey

Javascript

<script>

/* JavaScript program to check if
linked list is palindrome recursively */

    class Node {
        constructor(val) {
            this.data = val;
            this.ptr = null;
        }
    }

var one = new Node(1);
var two = new Node(2);
var three = new Node(3);
var four = new Node(4);
var five = new Node(3);
var six = new Node(2);
var seven = new Node(1);
        one.ptr = two;
        two.ptr = three;
        three.ptr = four;
        four.ptr = five;
        five.ptr = six;
        six.ptr = seven;
        var condition = isPalindrome(one);
        document.write("isPalidrome: " + condition);


    function isPalindrome(head) {

var slow = head;
        var ispalin = true;
        var stack = [];

        while (slow != null) {
            stack.push(slow.data);
            slow = slow.ptr;
        }

        while (head != null) {

            var i = stack.pop();
            if (head.data == i) {
                ispalin = true;
            } else {
                ispalin = false;
                break;
            }
            head = head.ptr;
        }
        return ispalin;
    }


// This code is contributed by todaysgaurav

</script>

Output

isPalindrome is true

Time complexity: O(N), Iterating over the linked list of size N.
Auxiliary Space: O(N), Using an auxiliary stack

Check if a Singly Linked List is Palindrome by Reversing the Linked List:

The idea is to first reverse the second half part of the linked list and then check whether the list is palindrome or not.

Follow the steps below to solve the problem:

• Get the middle of the linked list. 
• Reverse the second half of the linked list. 
• Check if the first half and second half are identical. 
• Construct the original linked list by reversing the second half again and attaching it back to the first half

Below is the implementation of the above approach:

C++

// C++ program to check if a linked list is palindrome
#include <bits/stdc++.h>
using namespace std;

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

void reverse(struct Node**);
bool compareLists(struct Node*, struct Node*);

// Function to check if given linked list is
// palindrome or not
bool isPalindrome(struct Node* head)
{
    struct Node *slow_ptr = head, *fast_ptr = head;
    struct Node *second_half, *prev_of_slow_ptr = head;

    // To handle odd size list
    struct Node* midnode = NULL;

    // initialize result
    bool res = true;

    if (head != NULL && head->next != NULL) {

        // Get the middle of the list. Move slow_ptr by 1
        // and fast_ptr by 2, slow_ptr will have the middle
        // node
        while (fast_ptr != NULL && fast_ptr->next != NULL) {
            fast_ptr = fast_ptr->next->next;

            // We need previous of the slow_ptr for
            // linked lists with odd elements
            prev_of_slow_ptr = slow_ptr;
            slow_ptr = slow_ptr->next;
        }

        // fast_ptr would become NULL when there
        // are even elements in list. And not NULL
        // for odd elements. We need to skip the
        // middle node for odd case and store it
        // somewhere so that we can restore the
        // original list
        if (fast_ptr != NULL) {
            midnode = slow_ptr;
            slow_ptr = slow_ptr->next;
        }

        // Now reverse the second half and
        // compare it with first half
        second_half = slow_ptr;

        // NULL terminate first half
        prev_of_slow_ptr->next = NULL;

        // Reverse the second half
        reverse(&second_half);

        // compare
        res = compareLists(head, second_half);

        // Construct the original list back
        reverse(
            &second_half); // Reverse the second half again

        // If there was a mid node (odd size case)
        // which was not part of either first half
        // or second half.
        if (midnode != NULL) {
            prev_of_slow_ptr->next = midnode;
            midnode->next = second_half;
        }
        else
            prev_of_slow_ptr->next = second_half;
    }
    return res;
}

// Function to reverse the linked list
// Note that this function may change
// the head
void reverse(struct Node** head_ref)
{
    struct Node* prev = NULL;
    struct Node* current = *head_ref;
    struct Node* next;

    while (current != NULL) {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;
    }
    *head_ref = prev;
}

// Function to check if two input
// lists have same data
bool compareLists(struct Node* head1, struct Node* head2)
{
    struct Node* temp1 = head1;
    struct Node* temp2 = head2;

    while (temp1 && temp2) {
        if (temp1->data == temp2->data) {
            temp1 = temp1->next;
            temp2 = temp2->next;
        }
        else
            return 0;
    }

    // Both are empty return 1
    if (temp1 == NULL && temp2 == NULL)
        return 1;

    // Will reach here when one is NULL
    // and other is not
    return 0;
}

// Push a node to linked list. Note
// that this function changes the head
void push(struct Node** head_ref, char new_data)
{

    // Allocate node
    struct Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));

    // Put in the data
    new_node->data = new_data;

    // Link the old list of the new node
    new_node->next = (*head_ref);

    // Move the head to point to the new node
    (*head_ref) = new_node;
}

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

// Driver code
int main()
{

    // Start with the empty list
    struct Node* head = NULL;
    char str[] = "abacaba";
    int i;

    for (i = 0; str[i] != '\0'; i++) {
        push(&head, str[i]);
    }
    isPalindrome(head) ? cout << "Is Palindrome"
                              << "\n\n"
                       : cout << "Not Palindrome"
                              << "\n\n";
    return 0;
}

// This code is contributed by Shivani

C

/*  C Program to check if a linked list is palindrome */
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>

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

void reverse(struct Node**);
bool compareLists(struct Node*, struct Node*);

/* Function to check if given linked list is
palindrome or not */
bool isPalindrome(struct Node* head)
{
    struct Node *slow_ptr = head, *fast_ptr = head;
    struct Node *second_half, *prev_of_slow_ptr = head;
    struct Node* midnode = NULL; // To handle odd size list
    bool res = true; // initialize result

    if (head != NULL && head->next != NULL) {
        /* Get the middle of the list. Move slow_ptr by 1
        and fast_ptr by 2, slow_ptr will have the middle
        node */
        while (fast_ptr != NULL && fast_ptr->next != NULL) {
            fast_ptr = fast_ptr->next->next;

            /*We need previous of the slow_ptr for
            linked lists with odd elements */
            prev_of_slow_ptr = slow_ptr;
            slow_ptr = slow_ptr->next;
        }

        /* fast_ptr would become NULL when there are even
        elements in list. And not NULL for odd elements. We
        need to skip the middle node for odd case and store
        it somewhere so that we can restore the original
        list*/
        if (fast_ptr != NULL) {
            midnode = slow_ptr;
            slow_ptr = slow_ptr->next;
        }

        // Now reverse the second half and compare it with
        // first half
        second_half = slow_ptr;
        prev_of_slow_ptr->next
            = NULL; // NULL terminate first half
        reverse(&second_half); // Reverse the second half
        res = compareLists(head, second_half); // compare

        /* Construct the original list back */
        reverse(
            &second_half); // Reverse the second half again

        // If there was a mid node (odd size case) which
        // was not part of either first half or second half.
        if (midnode != NULL) {
            prev_of_slow_ptr->next = midnode;
            midnode->next = second_half;
        }
        else
            prev_of_slow_ptr->next = second_half;
    }
    return res;
}

/* Function to reverse the linked list Note that this
    function may change the head */
void reverse(struct Node** head_ref)
{
    struct Node* prev = NULL;
    struct Node* current = *head_ref;
    struct Node* next;
    while (current != NULL) {
        next = current->next;
        current->next = prev;
        prev = current;
        current = next;
    }
    *head_ref = prev;
}

/* Function to check if two input lists have same data*/
bool compareLists(struct Node* head1, struct Node* head2)
{
    struct Node* temp1 = head1;
    struct Node* temp2 = head2;

    while (temp1 && temp2) {
        if (temp1->data == temp2->data) {
            temp1 = temp1->next;
            temp2 = temp2->next;
        }
        else
            return 0;
    }

    /* Both are empty return 1*/
    if (temp1 == NULL && temp2 == NULL)
        return 1;

    /* Will reach here when one is NULL
    and other is not */
    return 0;
}

/* Push a node to linked list. Note that this function
changes the head */
void push(struct Node** head_ref, char new_data)
{
    /* allocate node */
    struct Node* new_node
        = (struct Node*)malloc(sizeof(struct Node));

    /* put in the data */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

// A utility function to print a given linked list
void printList(struct Node* ptr)
{
    while (ptr != NULL) {
        printf("%c->", ptr->data);
        ptr = ptr->next;
    }
    printf("NULL\n");
}

/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    struct Node* head = NULL;
    char str[] = "abacaba";
    int i;

    for (i = 0; str[i] != '\0'; i++) {
        push(&head, str[i]);
    }
    isPalindrome(head) ? printf("Is Palindrome\n\n")
                       : printf("Not Palindrome\n\n");

    return 0;
}

Java

/* Java program to check if linked list is palindrome */

class LinkedList {
    Node head; // head of list
    Node slow_ptr, fast_ptr, second_half;

    /* Linked list Node*/
    class Node {
        char data;
        Node next;

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

    /* Function to check if given linked list is
       palindrome or not */
    boolean isPalindrome(Node head)
    {
        slow_ptr = head;
        fast_ptr = head;
        Node prev_of_slow_ptr = head;
        Node midnode = null; // To handle odd size list
        boolean res = true; // initialize result

        if (head != null && head.next != null) {
            /* Get the middle of the list. Move slow_ptr by
               1 and fast_ptr by 2, slow_ptr will have the
               middle node */
            while (fast_ptr != null
                   && fast_ptr.next != null) {
                fast_ptr = fast_ptr.next.next;

                /*We need previous of the slow_ptr for
                  linked lists  with odd elements */
                prev_of_slow_ptr = slow_ptr;
                slow_ptr = slow_ptr.next;
            }

            /* fast_ptr would become NULL when there are
               even elements in the list and not NULL for
               odd elements. We need to skip the middle node
               for odd case and store it somewhere so that
               we can restore the original list */
            if (fast_ptr != null) {
                midnode = slow_ptr;
                slow_ptr = slow_ptr.next;
            }

            // Now reverse the second half and compare it
            // with first half
            second_half = slow_ptr;
            prev_of_slow_ptr.next
                = null; // NULL terminate first half
            reverse(); // Reverse the second half
            res = compareLists(head,
                               second_half); // compare

            /* Construct the original list back */
            reverse(); // Reverse the second half again

            if (midnode != null) {
                // If there was a mid node (odd size case)
                // which was not part of either first half
                // or second half.
                prev_of_slow_ptr.next = midnode;
                midnode.next = second_half;
            }
            else
                prev_of_slow_ptr.next = second_half;
        }
        return res;
    }

    /* Function to reverse the linked list  Note that this
       function may change the head */
    void reverse()
    {
        Node prev = null;
        Node current = second_half;
        Node next;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        second_half = prev;
    }

    /* Function to check if two input lists have same data*/
    boolean compareLists(Node head1, Node head2)
    {
        Node temp1 = head1;
        Node temp2 = head2;

        while (temp1 != null && temp2 != null) {
            if (temp1.data == temp2.data) {
                temp1 = temp1.next;
                temp2 = temp2.next;
            }
            else
                return false;
        }

        /* Both are empty return 1*/
        if (temp1 == null && temp2 == null)
            return true;

        /* Will reach here when one is NULL
           and other is not */
        return false;
    }

    /* Push a node to linked list. Note that this function
       changes the head */
    public void push(char new_data)
    {
        /* Allocate the Node &
           Put in the data */
        Node new_node = new Node(new_data);

        /* link the old list of the new one */
        new_node.next = head;

        /* Move the head to point to new Node */
        head = new_node;
    }

    // A utility function to print a given linked list
    void printList(Node ptr)
    {
        while (ptr != null) {
            System.out.print(ptr.data + "->");
            ptr = ptr.next;
        }
        System.out.println("NULL");
    }

    /* Driver program to test the above functions */
    public static void main(String[] args)
    {

        /* Start with the empty list */
        LinkedList llist = new LinkedList();

        char str[] = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' };
        String string = new String(str);
        for (int i = 0; i < 7; i++) {
            llist.push(str[i]);
        }
        if (llist.isPalindrome(llist.head) != false) {
            System.out.println("Is Palindrome");
            System.out.println("");
        }
        else {
            System.out.println("Not Palindrome");
            System.out.println("");
        }
    }
}

Python3

# Python3 program to check if
# linked list is 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

    # Function to check if given
    # linked list is palindrome or not
    def isPalindrome(self, head):

        slow_ptr = head
        fast_ptr = head
        prev_of_slow_ptr = head

        # To handle odd size list
        midnode = None

        # Initialize result
        res = True

        if (head != None and head.next != None):

            # Get the middle of the list.
            # Move slow_ptr by 1 and
            # fast_ptr by 2, slow_ptr
            # will have the middle node
            while (fast_ptr != None and
                   fast_ptr.next != None):

                # We need previous of the slow_ptr
                # for linked lists  with odd
                # elements
                fast_ptr = fast_ptr.next.next
                prev_of_slow_ptr = slow_ptr
                slow_ptr = slow_ptr.next

            # fast_ptr would become NULL when
            # there are even elements in the
            # list and not NULL for odd elements.
            # We need to skip the middle node for
            # odd case and store it somewhere so
            # that we can restore the original list
            if (fast_ptr != None):
                midnode = slow_ptr
                slow_ptr = slow_ptr.next

            # Now reverse the second half
            # and compare it with first half
            second_half = slow_ptr

            # NULL terminate first half
            prev_of_slow_ptr.next = None

            # Reverse the second half
            second_half = self.reverse(second_half)

            # Compare
            res = self.compareLists(head, second_half)

            # Construct the original list back
            # Reverse the second half again
            second_half = self.reverse(second_half)

            if (midnode != None):

                # If there was a mid node (odd size
                # case) which was not part of either
                # first half or second half.
                prev_of_slow_ptr.next = midnode
                midnode.next = second_half
            else:
                prev_of_slow_ptr.next = second_half
        return res

    # Function to reverse the linked list
    # Note that this function may change
    # the head
    def reverse(self, second_half):

        prev = None
        current = second_half
        next = None

        while current != None:
            next = current.next
            current.next = prev
            prev = current
            current = next

        second_half = prev
        return second_half

    # Function to check if two input
    # lists have same data
    def compareLists(self, head1, head2):

        temp1 = head1
        temp2 = head2

        while (temp1 and temp2):
            if (temp1.data == temp2.data):
                temp1 = temp1.next
                temp2 = temp2.next
            else:
                return 0

        # Both are empty return 1
        if (temp1 == None and temp2 == None):
            return 1

        # Will reach here when one is NULL
        # and other is not
        return 0

    # Function to insert a new node
    # at the beginning
    def push(self, new_data):

        # Allocate the Node &
        # Put in the data
        new_node = Node(new_data)

        # Link the old list of the new one
        new_node.next = self.head

        # Move the head to point to new Node
        self.head = new_node

    # A utility function to print
    # a given linked list
    def printList(self):

        temp = self.head

        while(temp):
            print(temp.data, end="->")
            temp = temp.next

        print("NULL")


# Driver code
if __name__ == '__main__':

    l = LinkedList()
    s = ['a', 'b', 'a', 'c', 'a', 'b', 'a']

    for i in range(7):
        l.push(s[i])
    if (l.isPalindrome(l.head) != False):
        print("Is Palindrome\n")
    else:
        print("Not Palindrome\n")

# This code is contributed by MuskanKalra1

C#

/* C# program to check if linked list is palindrome */
using System;
class LinkedList {
    Node head; // head of list
    Node slow_ptr, fast_ptr, second_half;

    /* Linked list Node*/
    public class Node {
        public char data;
        public Node next;

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

    /* Function to check if given linked list is
       palindrome or not */
    Boolean isPalindrome(Node head)
    {
        slow_ptr = head;
        fast_ptr = head;
        Node prev_of_slow_ptr = head;
        Node midnode = null; // To handle odd size list
        Boolean res = true; // initialize result

        if (head != null && head.next != null) {
            /* Get the middle of the list. Move slow_ptr by
               1 and fast_ptr by 2, slow_ptr will have the
               middle node */
            while (fast_ptr != null
                   && fast_ptr.next != null) {
                fast_ptr = fast_ptr.next.next;

                /*We need previous of the slow_ptr for
                  linked lists  with odd elements */
                prev_of_slow_ptr = slow_ptr;
                slow_ptr = slow_ptr.next;
            }

            /* fast_ptr would become NULL when there are
               even elements in the list and not NULL for
               odd elements. We need to skip the middle node
               for odd case and store it somewhere so that
               we can restore the original list */
            if (fast_ptr != null) {
                midnode = slow_ptr;
                slow_ptr = slow_ptr.next;
            }

            // Now reverse the second half and compare it
            // with first half
            second_half = slow_ptr;
            prev_of_slow_ptr.next
                = null; // NULL terminate first half
            reverse(); // Reverse the second half
            res = compareLists(head,
                               second_half); // compare

            /* Construct the original list back */
            reverse(); // Reverse the second half again

            if (midnode != null) {
                // If there was a mid node (odd size case)
                // which was not part of either first half
                // or second half.
                prev_of_slow_ptr.next = midnode;
                midnode.next = second_half;
            }
            else
                prev_of_slow_ptr.next = second_half;
        }
        return res;
    }

    /* Function to reverse the linked list  Note that this
       function may change the head */
    void reverse()
    {
        Node prev = null;
        Node current = second_half;
        Node next;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        second_half = prev;
    }

    /* Function to check if two input lists have same data*/
    Boolean compareLists(Node head1, Node head2)
    {
        Node temp1 = head1;
        Node temp2 = head2;

        while (temp1 != null && temp2 != null) {
            if (temp1.data == temp2.data) {
                temp1 = temp1.next;
                temp2 = temp2.next;
            }
            else
                return false;
        }

        /* Both are empty return 1*/
        if (temp1 == null && temp2 == null)
            return true;

        /* Will reach here when one is NULL
           and other is not */
        return false;
    }

    /* Push a node to linked list. Note that this function
       changes the head */
    public void push(char new_data)
    {
        /* Allocate the Node &
           Put in the data */
        Node new_node = new Node(new_data);

        /* link the old list of the new one */
        new_node.next = head;

        /* Move the head to point to new Node */
        head = new_node;
    }

    // A utility function to print a given linked list
    void printList(Node ptr)
    {
        while (ptr != null) {
            Console.Write(ptr.data + "->");
            ptr = ptr.next;
        }
        Console.WriteLine("NULL");
    }

    /* Driver program to test the above functions */
    public static void Main(String[] args)
    {

        /* Start with the empty list */
        LinkedList llist = new LinkedList();

        char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' };

        for (int i = 0; i < 7; i++) {
            llist.push(str[i]);
        }
        if (llist.isPalindrome(llist.head) != false) {
            Console.WriteLine("Is Palindrome");
            Console.WriteLine("");
        }
        else {
            Console.WriteLine("Not Palindrome");
            Console.WriteLine("");
        }
    }
}
// This code is contributed by Arnab Kundu

Javascript

<script>
/* javascript program to check if linked list is palindrome */


    var head; // head of list
    var slow_ptr, fast_ptr, second_half;

    /* Linked list Node */
     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }

    /*
     * Function to check if given linked list is palindrome or not
     */
    function isPalindrome(head) {
        slow_ptr = head;
        fast_ptr = head;
var prev_of_slow_ptr = head;
var midnode = null; // To handle odd size list
        var res = true; // initialize result

        if (head != null && head.next != null) {
            /*
             * Get the middle of the list. Move slow_ptr by 1 and fast_ptr by 2, slow_ptr
             * will have the middle node
             */
            while (fast_ptr != null && fast_ptr.next != null) {
                fast_ptr = fast_ptr.next.next;

                /*
                 * We need previous of the slow_ptr for linked lists with odd elements
                 */
                prev_of_slow_ptr = slow_ptr;
                slow_ptr = slow_ptr.next;
            }

            /*
             * fast_ptr would become NULL when there are even elements in the list and not
             * NULL for odd elements. We need to skip the middle node for odd case and store
             * it somewhere so that we can restore the original list
             */
            if (fast_ptr != null) {
                midnode = slow_ptr;
                slow_ptr = slow_ptr.next;
            }

            // Now reverse the second half and compare it with first half
            second_half = slow_ptr;
            prev_of_slow_ptr.next = null; // NULL terminate first half
            reverse(); // Reverse the second half
            res = compareLists(head, second_half); // compare

            /* Construct the original list back */
            reverse(); // Reverse the second half again

            if (midnode != null) {
                // If there was a mid node (odd size case) which
                // was not part of either first half or second half.
                prev_of_slow_ptr.next = midnode;
                midnode.next = second_half;
            } else
                prev_of_slow_ptr.next = second_half;
        }
        return res;
    }

    /*
     * Function to reverse the linked list Note that this function may change the
     * head
     */
    function reverse() {
var prev = null;
var current = second_half;
var next;
        while (current != null) {
            next = current.next;
            current.next = prev;
            prev = current;
            current = next;
        }
        second_half = prev;
    }

    /* Function to check if two input lists have same data */
    function compareLists(head1,  head2) {
var temp1 = head1;
var temp2 = head2;

        while (temp1 != null && temp2 != null) {
            if (temp1.data == temp2.data) {
                temp1 = temp1.next;
                temp2 = temp2.next;
            } else
                return false;
        }

        /* Both are empty return 1 */
        if (temp1 == null && temp2 == null)
            return true;

        /*
         * Will reach here when one is NULL and other is not
         */
        return false;
    }

    /*
     * Push a node to linked list. Note that this function changes the head
     */
     function push( new_data) {
        /*
         * Allocate the Node & Put in the data
         */
var new_node = new Node(new_data);

        /* link the old list of the new one */
        new_node.next = head;

        /* Move the head to point to new Node */
        head = new_node;
    }

    // A utility function to print a given linked list
    function printList(ptr) {
        while (ptr != null) {
            document.write(ptr.data + "->");
            ptr = ptr.next;
        }
        document.write("NULL<br/>");
    }

    /* Driver program to test the above functions */


        /* Start with the empty list */

        var str = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ];
        var string = str.toString();
        for (i = 0; i < 7; i++) {
            push(str[i]);
            printList(head);
            if (isPalindrome(head) != false) {
                document.write("Is Palindrome");
                document.write("<br/>");
            } else {
                document.write("Not Palindrome");
                document.write("<br/>");
            }
        }


// This code contributed by gauravrajput1
</script>

Output

Is Palindrome

Time Complexity: O(N),  
Auxiliary Space: O(1)

Check if a Singly Linked List is Palindrome using Recursion:

The idea is to use the function call stack as a container. Recursively traverse till the end of the list. When returning from the last NULL, will be at the last node. The last node is to be compared with the first node of the list.

Follow the steps below to solve the problem:

• First make a recursive call to the next node of linked till it reach the last node.
• After returning from last node start checking from start of the linked list then move to the next node.
• Repeat these steps till reach the last node.

Below is the implementation of above approach:

C++

// Recursive program to check if a given linked list is
// palindrome
#include <bits/stdc++.h>
using namespace std;

/* Link list node */
struct node {
    char data;
    struct node* next;
};

// Initial parameters to this function are &head and head
bool isPalindromeUtil(struct node** left,
                      struct node* right)
{
    /* stop recursion when right becomes NULL */
    if (right == NULL)
        return true;

    /* If sub-list is not palindrome then no need to
    check for current left and right, return false */
    bool isp = isPalindromeUtil(left, right->next);
    if (isp == false)
        return false;

    /* Check values at current left and right */
    bool isp1 = (right->data == (*left)->data);

    /* Move left to next node */
    *left = (*left)->next;

    return isp1;
}

// A wrapper over isPalindromeUtil()
bool isPalindrome(struct node* head)
{
    return isPalindromeUtil(&head, head);
}

/* Push a node to linked list. Note that this function
changes the head */
void push(struct node** head_ref, char new_data)
{
    /* allocate node */
    struct node* new_node
        = (struct node*)malloc(sizeof(struct node));

    /* put in the data */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

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

/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    struct node* head = NULL;
    char str[] = "abacaba";
    int i;

    for (i = 0; str[i] != '\0'; i++) {
        push(&head, str[i]);
    }
    isPalindrome(head) ? cout << "Is Palindrome\n\n"
                       : cout << "Not Palindrome\n\n";
    return 0;
}
// this code is contributed by shivanisinghss2110

C

// Recursive program to check if a given linked list is
// palindrome
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>

/* Link list node */
struct node {
    char data;
    struct node* next;
};

// Initial parameters to this function are &head and head
bool isPalindromeUtil(struct node** left,
                      struct node* right)
{
    /* stop recursion when right becomes NULL */
    if (right == NULL)
        return true;

    /* If sub-list is not palindrome then no need to
    check for current left and right, return false */
    bool isp = isPalindromeUtil(left, right->next);
    if (isp == false)
        return false;

    /* Check values at current left and right */
    bool isp1 = (right->data == (*left)->data);

    /* Move left to next node */
    *left = (*left)->next;

    return isp1;
}

// A wrapper over isPalindromeUtil()
bool isPalindrome(struct node* head)
{
    return isPalindromeUtil(&head, head);
}

/* Push a node to linked list. Note that this function
changes the head */
void push(struct node** head_ref, char new_data)
{
    /* allocate node */
    struct node* new_node
        = (struct node*)malloc(sizeof(struct node));

    /* put in the data */
    new_node->data = new_data;

    /* link the old list of the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref) = new_node;
}

// A utility function to print a given linked list
void printList(struct node* ptr)
{
    while (ptr != NULL) {
        printf("%c->", ptr->data);
        ptr = ptr->next;
    }
    printf("NULL\n");
}

/* Driver program to test above function*/
int main()
{
    /* Start with the empty list */
    struct node* head = NULL;
    char str[] = "abacaba";
    int i;

    for (i = 0; str[i] != '\0'; i++) {
        push(&head, str[i]);
    }
    isPalindrome(head) ? printf("Is Palindrome\n\n")
                       : printf("Not Palindrome\n\n");
    return 0;
}

Java

// Java program for the above approach
public class LinkedList {

    // Head of the list
    Node head;
    Node left;

    public class Node {
        public char data;
        public Node next;

        // Linked list node
        public Node(char d)
        {
            data = d;
            next = null;
        }
    }

    // Initial parameters to this function are
    // &head and head
    boolean isPalindromeUtil(Node right)
    {
        left = head;

        // Stop recursion when right becomes null
        if (right == null)
            return true;

        // If sub-list is not palindrome then no need to
        // check for the current left and right, return
        // false
        boolean isp = isPalindromeUtil(right.next);
        if (isp == false)
            return false;

        // Check values at current left and right
        boolean isp1 = (right.data == left.data);

        left = left.next;

        // Move left to next node;
        return isp1;
    }

    // A wrapper over isPalindrome(Node head)
    boolean isPalindrome(Node head)
    {
        boolean result = isPalindromeUtil(head);
        return result;
    }

    // Push a node to linked list. Note that
    // this function changes the head
    public void push(char new_data)
    {

        // Allocate the node and put in the data
        Node new_node = new Node(new_data);

        // Link the old list of the new one
        new_node.next = head;

        // Move the head to point to new node
        head = new_node;
    }

    // A utility function to print a
    // given linked list
    void printList(Node ptr)
    {
        while (ptr != null) {
            System.out.print(ptr.data + "->");
            ptr = ptr.next;
        }
        System.out.println("Null");
    }

    // Driver Code
    public static void main(String[] args)
    {
        LinkedList llist = new LinkedList();
        char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' };
        for (int i = 0; i < 7; i++) {
            llist.push(str[i]);
        }
        if (llist.isPalindrome(llist.head)) {
            System.out.println("Is Palindrome");
            System.out.println("");
        }
        else {
            System.out.println("Not Palindrome");
            System.out.println("");
        }
    }
}

// This code is contributed by abhinavjain194

Python3

# Python program for the above approach

# Head of the list
head = None
left = None


class Node:
    def __init__(self, val):
        self.data = val
        self.next = None

# Initial parameters to this function are
# &head and head


def isPalindromeUtil(right):
    global head, left

    left = head

    # Stop recursion when right becomes null
    if (right == None):
        return True

    # If sub-list is not palindrome then no need to
    # check for the current left and right, return
    # false
    isp = isPalindromeUtil(right.next)
    if (isp == False):
        return False

    # Check values at current left and right
    isp1 = (right.data == left.data)

    left = left.next

    # Move left to next node;
    return isp1

# A wrapper over isPalindrome(Node head)


def isPalindrome(head):
    result = isPalindromeUtil(head)
    return result

# Push a node to linked list. Note that
# this function changes the head


def push(new_data):
    global head

    # Allocate the node and put in the data
    new_node = Node(new_data)

    # Link the old list of the new one
    new_node.next = head

    # Move the head to point to new node
    head = new_node

# A utility function to print a
# given linked list


def printList(ptr):
    while (ptr != None):
        print(ptr.data, end="->")
        ptr = ptr.next

    print("Null ")


# Driver Code
str = ['a', 'b', 'a', 'c', 'a', 'b', 'a']

for i in range(0, 7):
    push(str[i])

if (isPalindrome(head) and i != 0):
    print("Is Palindrome\n")
else:
    print("Not Palindrome\n")

# This code is contributed by saurabh_jaiswal.

C#

/* C# program to check if linked list
is palindrome recursively */
using System;

public class LinkedList {
    Node head; // head of list
    Node left;

    /* Linked list Node*/
    public class Node {
        public char data;
        public Node next;

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

    // Initial parameters to this function are &head and
    // head
    Boolean isPalindromeUtil(Node right)
    {
        left = head;

        /* stop recursion when right becomes NULL */
        if (right == null)
            return true;

        /* If sub-list is not palindrome then no need to
        check for current left and right, return false */
        Boolean isp = isPalindromeUtil(right.next);
        if (isp == false)
            return false;

        /* Check values at current left and right */
        Boolean isp1 = (right.data == (left).data);

        /* Move left to next node */
        left = left.next;

        return isp1;
    }

    // A wrapper over isPalindromeUtil()
    Boolean isPalindrome(Node head)
    {
        Boolean result = isPalindromeUtil(head);
        return result;
    }

    /* Push a node to linked list. Note that this function
    changes the head */
    public void push(char new_data)
    {
        /* Allocate the Node &
        Put in the data */
        Node new_node = new Node(new_data);

        /* link the old list of the new one */
        new_node.next = head;

        /* Move the head to point to new Node */
        head = new_node;
    }

    // A utility function to print a given linked list
    void printList(Node ptr)
    {
        while (ptr != null) {
            Console.Write(ptr.data + "->");
            ptr = ptr.next;
        }
        Console.WriteLine("NULL");
    }

    /* Driver code */
    public static void Main(String[] args)
    {
        /* Start with the empty list */
        LinkedList llist = new LinkedList();

        char[] str = { 'a', 'b', 'a', 'c', 'a', 'b', 'a' };
        // String string = new String(str);
        for (int i = 0; i < 7; i++) {
            llist.push(str[i]);
        }

        if (llist.isPalindrome(llist.head) != false) {
            Console.WriteLine("Is Palindrome");
            Console.WriteLine("");
        }
        else {
            Console.WriteLine("Not Palindrome");
            Console.WriteLine("");
        }
    }
}

// This code is contributed by Rajput-Ji

Javascript

<script>
// javascript program for the above approach


    // Head of the list
    var head;
    var left;

     class Node {
            constructor(val) {
                this.data = val;
                this.next = null;
            }
        }

    // Initial parameters to this function are
    // &head and head
    function isPalindromeUtil( right) {
        left = head;

        // Stop recursion when right becomes null
        if (right == null)
            return true;

        // If sub-list is not palindrome then no need to
        // check for the current left and right, return
        // false
        var isp = isPalindromeUtil(right.next);
        if (isp == false)
            return false;

        // Check values at current left and right
        var isp1 = (right.data == left.data);

        left = left.next;

        // Move left to next node;
        return isp1;
    }

    // A wrapper over isPalindrome(Node head)
    function isPalindrome( head) {
        var result = isPalindromeUtil(head);
        return result;
    }

    // Push a node to linked list. Note that
    // this function changes the head
    function push( new_data) {

        // Allocate the node and put in the data
        var new_node = new Node(new_data);

        // Link the old list of the new one
        new_node.next = head;

        // Move the head to point to new node
        head = new_node;
    }

    // A utility function to print a
    // given linked list
    function printList( ptr) {
        while (ptr != null) {
            document.write(ptr.data + "->");
            ptr = ptr.next;
        }
        document.write("Null ");
        document.write("<br>");

    }

    // Driver Code

        var str = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ];
        for (var i = 0; i < 7; i++) {
            push(str[i]);
            printList(head);

            if (isPalindrome(head)) {
                document.write("Is Palindrome");
                document.write("<br/>");
                document.write("<br>");
            } else {
                document.write("Not Palindrome");
                document.write("<br/>");
                document.write("<br/>");
            }
        }

// This code contributed by aashish1995

</script>

Output

Is Palindrome

Time Complexity: O(N), Traversing over the linked list of size N.
Auxiliary Space: O(N) if Function Call Stack size is considered, otherwise O(1).

Thanks to Sharad Chandra for suggesting this approach.