### Problem Statement

We will be given a linked list, and we need to find whether the given list is palindrome or not.

### Problem Statement Understanding

To understand this problem statement, let us take examples.

If the given linked list is 3â†’1â†’18â†’1â†’3â†’NULL, and now according to the problem statement, we need to determine whether the given linked list is palindrome or not.

- If we iterate the given list forward and print the elements, we will get: 3,1,18,1,3
- Now, if we iterate the given list backward and print the elements, we will get: 3,1,18,1,3.
- We can see that no matter we iterate forward or backward, we are getting the same output at the end. So, the given list is a palindrome.

At this point, we have understood the problem statement. Now we will try to formulate an approach for this problem.

Before moving to the approach section, try to think about how you can approach this problem.

- If stuck, no problem, we will thoroughly see how we can approach this problem in the next section.

Letâ€™s move to the approach section.

### Approach 1

To check whether a linked list is a palindrome or not, we need to traverse the list backward from the last node of the list, but as our linked list is a singly linked list, we cannot move backwards in the list.

To solve the above problem, we will use a stack and store all the elements of the list in the stack while forward iteration of the list.

- After that, we will again iterate the list and will simultaneously pop an element from the stack at each iteration and will check if this popped element is equal to the current element of the list or not.
- If all the elements are equal, then the list is a palindrome. Else, It is not a palindrome.

**Time Complexity:** O(n), where n is the total number of nodes in the list

**Space Complexity:** O(n), due to the stack

The above approach seems fine, but we are using a stack that is increasing the space complexity. So, we need to think of a better approach where no extra space is used.

### Approach 2

Since we cannot move backward in the list, we need to devise an algorithm to get our work done.

We can reverse the second half of the list and then simultaneously compare the elements from the first and second half of the list.

- If all elements of first and the second half are equal, then the list is a palindrome.
- Else, the list is not a palindrome.

### Algorithm

1) If the list is empty or it has only one node then simply return true as an empty list or a list with a single node is always a palindrome.

2) Initialize two pointers **slow** and **fast** with the first node of the list (finding the middle node of the linked list).

- Run a while loop till the
**fast**reach the last or last second node of the list. - In each iteration, advance
**slow**forward by one node and**fast**forward by two nodes. - Now, when the
**fast**is at the end of the list,**slow**will be pointing to the middle node of the linked list.

3) Now, reverse the second half of the linked list starting from**slow->next**to the last node of the linked list.

4) After that, simultaneously iterate through the first half and the second half of the linked list using pointers**head**and**slow**: - If the data of the node to which
**head**and**slow**are pointing are not equal, return false. - Else, advance
**slow**and**head**by one node.

5) After execution of the above while loop, return true from the function.

### Dry Run

### Code Implementation

#include <stdio.h> #include <stdbool.h> //Represent a node of the singly linked list struct node{ int data; struct node *next; }; //Represent the head and tail of the singly linked list struct node *head, *tail = NULL; int size = 0; //addNode() will add a new node to the list void addNode(int data) { //Create a new node struct node *newNode = (struct node*)malloc(sizeof(struct node)); newNode->data = data; newNode->next = NULL; //Checks if the list is empty if(head == NULL) { //If list is empty, both head and tail will point to new node head = newNode; tail = newNode; } else { //newNode will be added after tail such that tail's next will point to newNode tail->next = newNode; //newNode will become new tail of the list tail = newNode; } //Size will count the number of nodes present in the list size++; } //reverseList() will reverse the singly linked list and return the head of the list struct node* reverseList(struct node *temp){ struct node *current = temp; struct node *prevNode = NULL, *nextNode = NULL; //Swap the previous and next nodes of each node to reverse the direction of the list while(current != NULL){ nextNode = current->next; current->next = prevNode; prevNode = current; current = nextNode; } return prevNode; } //isPalindrome() will determine whether given list is palindrome or not. void isPalindrome(){ struct node *current = head; bool flag = true; //Store the mid position of the list int mid = (size%2 == 0)? (size/2) : ((size+1)/2); //Finds the middle node in given singly linked list for(int i=1; i<mid; i++){="" current="current-">next; } //Reverse the list after middle node to end struct node *revHead = reverseList(current->next); //Compare nodes of first half and second half of list while(head != NULL && revHead != NULL){ if(head->data != revHead->data){ flag = false; break; } head = head->next; revHead = revHead->next; } if(flag) printf("Given singly linked list is a palindrome\n"); else printf("Given singly linked list is not a palindrome\n"); } //display() will display all the nodes present in the list void display() { //Node current will point to head struct node *current = head; if(head == NULL) { printf("List is empty\n"); return; } printf("Nodes of singly linked list: \n"); while(current != NULL) { //Prints each node by incrementing pointer printf("%d ", current->data); current = current->next; } printf("\n"); } int main() { //Add nodes to the list addNode(1); addNode(2); addNode(3); addNode(2); addNode(2); display(); //Checks whether given list is palindrome or not isPalindrome(); return 0; }

#include<bits stdc++.h=""> using namespace std; class Node{ public: int data; Node* next; Node(int x){ data = x; next = NULL; } }; // This function will reverse the list Node* reverse(Node *head) { Node *prev = NULL, *current = head, *next; while (current != NULL) { next = current->next; current->next = prev; prev = current; current = next; } return prev; } // This function checks if a list is palindrome bool isPalindrome(Node* head) { //check if the list is empty or //it has a single node if(head==NULL||head->next==NULL) return true; //initialize two pointers Node* slow=head; Node* fast=head; // find the middle node of the list while(fast->next!=NULL&&fast->next->next!=NULL){ slow=slow->next; fast=fast->next->next; } // reverse the second half of the list slow->next=reverse(slow->next); //update the 'slow' pointer slow=slow->next; //iterate both the lists simultaneously while(slow!=NULL){ if(head->data!=slow->data) return false; head=head->next; slow=slow->next; } return true; } int main(void){ Node* head = NULL; head = new Node(3); head->next = new Node(1); head->next->next = new Node(18); head->next->next->next = new Node(1); head->next->next->next->next = new Node(3); if(isPalindrome(head)){ cout<<"The list is a palindrome"; }else{ cout<<"The list is not a palindrome"; } return 0; }

class Node{ int data; Node next; Node(int x){ data = x; } } public class Palindrome { // This function will reverse the list static Node reverse(Node head) { Node prev =null, current = head, next=null; while (current != null) { next = current.next; current.next = prev; prev = current; current = next; } return prev; } // This function checks if a list is palindrome static boolean isPalindrome(Node head) { //check if the list is empty or //it has a single node if(head==null||head.next==null) return true; //initialize two pointers Node slow=head; Node fast=head; // find the middle node of the list while(fast.next!=null && fast.next.next!=null){ slow=slow.next; fast=fast.next.next; } // reverse the second half of the list slow.next=reverse(slow.next); //update the 'slow' pointer slow=slow.next; //iterate both the lists simultaneously while(slow!=null){ if(head.data!=slow.data) return false; head=head.next; slow=slow.next; } return true; } public static void main(String[] args) { Node head = null; head = new Node(3); head.next = new Node(1); head.next.next = new Node(18); head.next.next.next = new Node(1); head.next.next.next.next = new Node(3); if(isPalindrome(head)){ System.out.println("List is palindrome"); }else{ System.out.println("List is not palindrome"); } } }

class Node: def __init__(self,data): self.data = data self.next = None # This function will reverse the list def reverse(head): prev = None current = head while(current is not None): next = current.next current.next = prev prev = current current = next head = prev return head # This function checks if a list is palindrome def ispalindrome(head): # check if the list is empty or it has a single node if head == None or head.next == None: return True # initialize two pointers slow = head fast = head # find the middle node of the list while fast.next and fast.next.next: slow = slow.next fast = fast.next.next # reverse the second half of the list slow.next = reverse(slow.next) # update the 'slow' pointer slow = slow.next # iterate both the lists simultaneously while slow: if head.data != slow.data: return False head = head.next slow = slow.next return True head = None head = Node(3) head.next = Node(1) head.next.next = Node(18) head.next.next.next = Node(1) head.next.next.next.next = Node(3) if ispalindrome(head): print("The list is a palindrome") else: print("The list is not a palindrome")

#### Output

The list is a palindrome

**Time Complexity:** O(n), n is the total number of nodes in the list.

[forminator_quiz id=”5808″]

So, in this blog, we have tried to explain how you can Check If a Linked List Is Palindrome most optimally. If you want to solve more questions on Linked List, which are curated by our expert mentors at PrepBytes, you can follow this link Linked List.