Function to check if a singly linked list is palindrome

24 Jun, 2022
Given a singly linked list of characters, write a function that returns true if the given list is a palindrome, else false.
METHOD 1 (Use a Stack) 
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 image is a dry run of the above approach: 
Below is the implementation of the above approach : 
C++
Java
Python3
C#
Javascript
#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 true\n";  return 0;}  // This code has been contributed by Striver 
/* 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 program to check if linked# list is palindrome using stackclass Node:    def __init__(self,data):                  self.data = data        self.ptr = None          # Function to check if the linked list# is palindrome or notdef 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 listone = 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 pointerone.ptr = twotwo.ptr = threethree.ptr = fourfour.ptr = fivefive.ptr = sixsix.ptr = sevenseven.ptr = None  # Call function to check palindrome or notresult = ispalindrome(one)  print("isPalindrome:", result)  # This code is contributed by Nishtha Goel
// C# program to check if linked list// is palindrome recursively using System;using System.Collections.Generic;  class linkedList{      // Driver codepublic 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
<script>  /* JavaScript program to check iflinked 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: true
Time complexity: O(n)
Auxiliary Space: O (n) since we are using an auxiliary stack
METHOD 2 (By reversing the list) This method takes O(n) time and O(1) extra space. 1) Get the middle of the linked list. 2) Reverse the second half of the linked list. 3) Check if the first half and second half are identical. 4) Construct the original linked list by reversing the second half again and attaching it back to the first half
To divide the list into two halves, method 2 of this post is used. 
When a number of nodes are even, the first and second half contain exactly half nodes. The challenging thing in this method is to handle the case when the number of nodes is odd. We don’t want the middle node as part of the lists as we are going to compare them for equality. For odd cases, we use a separate variable ‘midnode’. 
C++
C
Java
Python3
C#
Javascript
// 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 databool 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 off 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]);         printList(head);         isPalindrome(head) ? cout << "Is Palindrome"                  << "\n\n" : cout << "Not Palindrome"                  << "\n\n";     }     return 0; }   // This code is contributed by Shivani
/* 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 off the new node */    new_node->next = (*head_ref);       /* move the head to pochar 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]);         printList(head);         isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n");     }       return 0; } 
/* 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 off 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]);            llist.printList(llist.head);            if (llist.isPalindrome(llist.head) != false) {                System.out.println("Is Palindrome");                System.out.println("");            }            else {                System.out.println("Not Palindrome");                System.out.println("");            }        }    }}
# Python3 program to check if# linked list is palindrome  # Node classclass 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 off 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 codeif __name__ == '__main__':          l = LinkedList()    s = [ 'a', 'b', 'a', 'c', 'a', 'b', 'a' ]          for i in range(7):        l.push(s[i])        l.printList()                  if (l.isPalindrome(l.head) != False):            print("Is Palindrome\n")        else:            print("Not Palindrome\n")        print()  # This code is contributed by MuskanKalra1
/* 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 off 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]);            llist.printList(llist.head);            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
<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 off 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: 
a->NULL
Is Palindrome

b->a->NULL
Not Palindrome

a->b->a->NULL
Is Palindrome

c->a->b->a->NULL
Not Palindrome

a->c->a->b->a->NULL
Not Palindrome

b->a->c->a->b->a->NULL
Not Palindrome

a->b->a->c->a->b->a->NULL
Is Palindrome
Time Complexity: O(n) Auxiliary Space: O(1)  
METHOD 3 (Using Recursion) Use two pointers left and right. Move right and left using recursion and check for following in each recursive call. 1) Sub-list is a palindrome. 2) Value at current left and right are matching.
If both above conditions are true then return true.
The idea is to use function call stack as a container. Recursively traverse till the end of list. When we return from last NULL, we will be at the last node. The last node to be compared with first node of list.
In order to access first node of list, we need list head to be available in the last call of recursion. Hence, we pass head also to the recursive function. If they both match we need to compare (2, n-2) nodes. Again when recursion falls back to (n-2)nd node, we need reference to 2nd node from the head. We advance the head pointer in the previous call, to refer to the next node in the list.However, the trick is identifying a double-pointer. Passing a single pointer is as good as pass-by-value, and we will pass the same pointer again and again. We need to pass the address of the head pointer for reflecting the changes in parent recursive calls.Thanks to Sharad Chandra for suggesting this approach.  
C++
C
Java
Python3
C#
Javascript
// 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) {     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 off 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]);         printList(head);         isPalindrome(head) ? cout << "Is Palindrome\n\n" : cout << "Not Palindrome\n\n";     }       return 0; } //this code is contributed by shivanisinghss2110 
// 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) {     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 off the new node */    new_node->next = (*head_ref);       /* move the head to pochar 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]);         printList(head);         isPalindrome(head) ? printf("Is Palindrome\n\n") : printf("Not Palindrome\n\n");     }       return 0; } 
// Java program for the above approachpublic class LinkedList{      // Head of the listNode 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 headboolean 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 headpublic void push(char new_data){          // Allocate the node and put in the data    Node new_node = new Node(new_data);      // Link the old list off the 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 listvoid printList(Node ptr){    while (ptr != null)     {        System.out.print(ptr.data + "->");        ptr = ptr.next;    }    System.out.println("Null");}  // Driver Codepublic 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]);        llist.printList(llist.head);                  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
# Python program for the above approach  # Head of the listhead = Noneleft = None  class Node:    def __init__(self, val):        self.data = val        self.next = None  # Initial parameters to this function are# &head and headdef 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 headdef push(new_data):    global head      # Allocate the node and put in the data    new_node = Node(new_data)      # Link the old list off the 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 listdef printList(ptr):    while (ptr != None):        print(ptr.data, end="->")        ptr = ptr.next      print("Null ")    # Driver Codestr = ['a', 'b', 'a', 'c', 'a', 'b', 'a']  for i in range(0, 7):    push(str[i])    printList(head)      if (isPalindrome(head) and i != 0):        print("Is Palindrome\n")    else:        print("Not Palindrome\n")  # This code is contributed by saurabh_jaiswal.
/* 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 off 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]);             llist.printList(llist.head);             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
<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 off the 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: 
a->NULL
Not Palindrome

b->a->NULL
Not Palindrome

a->b->a->NULL
Is Palindrome

c->a->b->a->NULL
Not Palindrome

a->c->a->b->a->NULL
Not Palindrome

b->a->c->a->b->a->NULL
Not Palindrome

a->b->a->c->a->b->a->NULL
Is Palindrome
Time Complexity: O(n) Auxiliary Space: O(n) if Function Call Stack size is considered, otherwise O(1). 
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palindrome
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