Last Updated on November 28, 2022 by Prepbytes
In this blog, we have to figure out a palindrome linked list. A Palindrome linked list means a Linked List which has values in a palindromic form i.e. values from left to right will be the same as the values from right to left in the linked list. Let’s discuss how to find a palindrome linked list.
How to check Palindrome Linked List
Check whether the given linked list is a palindrome or not.
Example:
2
3
2 5 2 true
5
4 5 6 3 4 false
First list is a palindrome since the first element is same as the last and middle one is common.
Second list is not a palindrome because the second element does not match the fourth element in the list.See original problem statement here
EXPLANATION:
Approach 1(Linked list reversal) For Palindrome Linked List:
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.
Note: This approach takes O(n) time and O(1) extra space.
Approach 2(Using recursion) For Palindrome Linked List:
Use two pointers
leftandright. Move right and left using recursion and check for following in each recursive call.Sub-list is palindrome.Value at current left and right are matching.
If both above conditions are true then return true.
bool isPalindrome(Node*& left, Node* right)
{
// base case
if (right == nullptr)
return true;
// return false on first mismatch
if (!isPalindrome(left, right->next))
return false;
// copy left pointer
Node* prevLeft = left;
// advance the left pointer to the next node
// this change would reflect in the parent recursive calls
left = left->next;
// In order for linked list to be palindrome, the character at the left
// node should match with the character at the right node
return (prevLeft->data == right->data);
}
Approach 3(Using stack) For Palindrome Linked List:
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 nodes with currently visited node.If all nodes matched, then return true, else false.
Note: This approach takes O(n) time and O(n) extra space(stack).
SOLUTIONS For Palindrome Linked List:
#include <stdio.h>
#include <stdbool.h>
struct node{
int data;
struct node *next;
};
struct node *head, *tail = NULL;
int size = 0;
void addNode(int data) {
struct node *newNode = (struct node*)malloc(sizeof(struct node));
newNode->data = data;
newNode->next = NULL;
if(head == NULL) {
head = newNode;
tail = newNode;
}
else {
tail->next = newNode;
tail = newNode;
}
size++;
}
struct node* reverseList(struct node *temp){
struct node *current = temp;
struct node *prevNode = NULL, *nextNode = NULL;
while(current != NULL){
nextNode = current->next;
current->next = prevNode;
prevNode = current;
current = nextNode;
}
return prevNode;
}
void isPalindrome(){
struct node *current = head;
bool flag = true;
int mid = (size%2 == 0)? (size/2)-1 : ((size)/2);
for(int i=1; i<mid; i++){
current = current->next;
}
struct node *revHead = reverseList(current->next);
while(head != NULL && revHead != NULL){
if(head->data != revHead->data){
flag = false;
break;
}
head = head->next;
revHead = revHead->next;
}
if(flag)
printf("true\n");
else
printf("false\n");
}
int main()
{
//Add nodes to the list
int t;
scanf("%d",&t);
while(t--)
{
int n;scanf("%d",&n);
for(int i=0;i<n;i++)
{
int x;scanf("%d",&x);
addNode(x);
}size=n;
isPalindrome();
}
return 0;
}
#include<bits/stdc++.h>
using namespace std;
struct node{
int data;
struct node *next;
};
struct node *head, *tail = NULL;
int size = 0;
void addNode(int data) {
node *newNode = new node();
newNode->data = data;
newNode->next = NULL;
if(head == NULL) {
head = newNode;
tail = newNode;
}
else {
tail->next = newNode;
tail = newNode;
}
size++;
}
struct node* reverseList(struct node *temp){
struct node *current = temp;
struct node *prevNode = NULL, *nextNode = NULL;
while(current != NULL){
nextNode = current->next;
current->next = prevNode;
prevNode = current;
current = nextNode;
}
return prevNode;
}
void isPalindrome(){
struct node *current = head;
bool flag = true;
int mid = (size%2 == 0)? (size/2)-1 : ((size)/2);
for(int i=1; i<mid; i++){
current = current->next;
}
struct node *revHead = reverseList(current->next);
while(head != NULL && revHead != NULL){
if(head->data != revHead->data){
flag = false;
break;
}
head = head->next;
revHead = revHead->next;
}
if(flag)
printf("true\n");
else
printf("false\n");
}
int main()
{
//Add nodes to the list
int t;
cin>>t;
while(t--)
{
int n;cin>>n;
for(int i=0;i<n;i++)
{
int x;cin>>x;
addNode(x);
}size=n;
isPalindrome();
}
return 0;
}
import java.io.*;
import java.util.*;
public class Main{
static class SinglyLinkedListNode {
public int data;
public SinglyLinkedListNode next;
public SinglyLinkedListNode(int nodeData) {
this.data = nodeData;
this.next = null;
}
}
static class SinglyLinkedList {
public SinglyLinkedListNode head;
public SinglyLinkedListNode tail;
public SinglyLinkedList() {
this.head = null;
this.tail = null;
}
public void insertNode(int nodeData) {
SinglyLinkedListNode node = new SinglyLinkedListNode(nodeData);
if (this.head == null) {
this.head = node;
} else {
this.tail.next = node;
}
this.tail = node;
}
}
static boolean palindromeLlist(SinglyLinkedListNode head) {
SinglyLinkedListNode slow = head;
boolean ispalin = true;
Stack<Integer> stack = new Stack<Integer>();
while (slow != null) {
stack.push(slow.data);
slow = slow.next;
}
while (head != null) {
int i = stack.pop();
if (head.data == i) {
ispalin = true;
}
else {
ispalin = false;
break;
}
head = head.next;
}
return ispalin;
}
private static final Scanner scanner = new Scanner(System.in);
public static void main(String[] args) throws IOException {
int testCases = scanner.nextInt();
while (testCases-- > 0) {
SinglyLinkedList llist = new SinglyLinkedList();
int llistCount = scanner.nextInt();
for (int i = 0; i < llistCount; i++) {
int llistItem = scanner.nextInt();
llist.insertNode(llistItem);
}
boolean res =palindromeLlist(llist.head);
if(res)
System.out.println("true");
else
System.out.println("false");
}
scanner.close();
}
}
class Node:
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
def __init__(self):
self.head = None
def isPalindrome(self, head):
slow_ptr = head
fast_ptr = head
prev_of_slow_ptr = head
midnode = None
res = True
if (head != None and head.next != None):
while (fast_ptr != None and
fast_ptr.next != None):
fast_ptr = fast_ptr.next.next
prev_of_slow_ptr = slow_ptr
slow_ptr = slow_ptr.next
if (fast_ptr != None):
midnode = slow_ptr
slow_ptr = slow_ptr.next
second_half = slow_ptr
prev_of_slow_ptr.next = None
second_half = self.reverse(second_half)
res = self.compareLists(head, second_half)
second_half = self.reverse(second_half)
if (midnode != None):
prev_of_slow_ptr.next = midnode
midnode.next = second_half
else:
prev_of_slow_ptr.next = second_half
return res
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
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
if (temp1 == None and temp2 == None):
return 1
return 0
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
def printList(self):
temp = self.head
while(temp):
print(temp.data, end = "->")
temp = temp.next
print("NULL")
if __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 article tried to discuss different approaches for finding palindrome linked list by which topics like stack, recursion and linked list are improved. Having knowledge about the topics like stack, recursion, and linked list will give an upper hand in the technical interviews. To practice more problems on Linked List, Recursion, Stack you can check out MYCODE | Competitive Programming.
FAQ
1. Can linked list be palindromes?
To check if a linked list is palindrome or not, we need to compare the first element with the last element, the second element with the second last element, the third element with the third last element, etc. If all the comparisons are equal, then the linked list is palindrome; otherwise, not.
2. Which companies asked in their interview for finding palindromic linked list?
IBM, Mcafee, Infosys, TCS and Gemini Solutions have recently asked this question in their interview.