Program To Determine Whether A Singly Linked List Is The Palindrome

Program To Determine Whether A Singly Linked List Is The Palindrome on fibonacci, factorial, prime, armstrong, swap, reverse, search, sort, stack, queue, array, linkedlist, tree, graph etc.

Program to determine whether a singly linked list is the palindrome

Explanation

In this program, we need to check whether given singly linked list is a palindrome or not. A palindromic list is the one which is equivalent to the reverse of itself.

The list given in the above figure is a palindrome since it is equivalent to its reverse list, i.e., 1, 2, 3, 2, 1. To check whether a list is a palindrome, we traverse the list and check if any element from the starting half doesn't match with any element from the ending half, then we set the variable flag to false and break the loop.

In the last, if the flag is false, then the list is palindrome otherwise not. The algorithm to check whether a list is a palindrome or not is given below.

Algorithm

1. Create a class Node which has two attributes: data and next. Next is a pointer to the next node in the list.
2. Create another class Palindrome which has three attributes: head, tail, and size.
1. Create a new node.
2. It first checks, whether the head is equal to null which means the list is empty.
3. If the list is empty, both head and tail will point to a newly added node.
4. If the list is not empty, the new node will be added to end of the list such that tail's next will point to a newly added node. This new node will become the new tail of the list.
4. reverseList() will reverse the order of the node present in the list:
1. Node current will represent a node from which a list needs to be reversed.
2. Node prevNode represent the previous node to current and nextNode represent the node next to current.
3. The list will be reversed by swapping the prevNode with nextNode for each node.
5. isPalindrome() will check whether given list is palindrome or not:
1. Declare a node current which will initially point to head node.
2. The variable flag will store a boolean value true.
3. Calculate the mid-point of the list by dividing the size of the list by 2.
4. Traverse through the list till current points to the middle node.
5. Reverse the list after the middle node until the last node using reverseList(). This list will be the second half of the list.
6. Now, compare nodes of first half and second half of the list.
7. If any of the nodes don't match then, set a flag to false and break the loop.
8. If the flag is true after the loop which denotes that list is a palindrome.
9. If the flag is false, then the list is not a palindrome.
6. display() will display the nodes present in the list:
1. Define a node current which will initially point to the head of the list.
2. Traverse through the list till current points to null.
3. Display each node by making current to point to node next to it in each iteration.

Python

```#Represent a node of the singly linked list
class Node:
def __init__(self,data):
self.data = data;
self.next = None;

class Palindrome:
def __init__(self):
self.tail = None;
self.size = 0;

#Create a new node
newNode = Node(data);

#Checks if the list is empty
#If list is empty, both head and tail will point to new node
self.tail = newNode;
else:
#newNode will be added after tail such that tail's next will point to newNode
self.tail.next = newNode;
#newNode will become new tail of the list
self.tail = newNode;
#Size will count the number of nodes present in the list
self.size = self.size + 1;

#reverseList() will reverse the singly linked list and return the head of the list
def reverseList(self, temp):
current = temp;
prevNode = None;
nextNode = None;

#Swap the previous and next nodes of each node to reverse the direction of the list
while(current != None):
nextNode = current.next;
current.next = prevNode;
prevNode = current;
current = nextNode;
return prevNode;

#isPalindrome() will determine whether given list is palindrome or not.
def isPalindrome(self):
flag = True;

#Store the mid position of the list
mid = (self.size//2) if(self.size%2 == 0) else ((self.size+1)//2);

#Finds the middle node in given singly linked list
for i in range(1, mid):
current = current.next;

#Reverse the list after middle node to end

#Compare nodes of first half and second half of list
flag = False;
break;

if(flag):
print("Given singly linked list is a palindrome");
else:
print("Given singly linked list is not a palindrome");

#display() will display all the nodes present in the list
def display(self):
#Node current will point to head

print("List is empty");
return;
print("Nodes of singly linked list: ");
while(current != None):
#Prints each node by incrementing pointer
print(current.data , end=" ");
current = current.next;
print();

sList = Palindrome();

sList.display();

#Checks whether given list is palindrome or not
sList.isPalindrome();
```

Output:

``` Nodes of the singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
```

C

```#include <stdio.h>
#include <stdbool.h>

//Represent a node of the singly linked list
struct node{
int data;
struct node *next;
};

struct node *head, *tail = NULL;
int size = 0;

//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 list is empty, both head and tail will point to new node
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(){
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

//Compare nodes of first half and second half of list
flag = false;
break;
}
}

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

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()
{

display();

//Checks whether given list is palindrome or not
isPalindrome();

return 0;
}
```

Output:

```Nodes of the singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
```

JAVA

```public class Palindrome {

//Represent a node of the singly linked list
class Node{
int data;
Node next;

public Node(int data) {
this.data = data;
this.next = null;
}
}

public int size;
public Node tail = null;

//Create a new node
Node newNode = new Node(data);

//Checks if the list is empty
//If list is empty, both head and tail will point to new node
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
public Node reverseList(Node temp){
Node current = temp;
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.
public void isPalindrome(){
boolean 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

//Compare nodes of first half and second half of list
flag = false;
break;
}
}

if(flag)
System.out.println("Given singly linked list is a palindrome");
else
System.out.println("Given singly linked list is not a palindrome");
}

//display() will display all the nodes present in the list
public void display() {
//Node current will point to head

System.out.println("List is empty");
return;
}
System.out.println("Nodes of singly linked list: ");
while(current != null) {
//Prints each node by incrementing pointer
System.out.print(current.data + " ");
current = current.next;
}
System.out.println();
}

public static void main(String[] args) {

Palindrome sList = new Palindrome();

sList.display();

//Checks whether given list is palindrome or not
sList.isPalindrome();
}
}
```

Output:

```Nodes of singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
```

C#

```using System;

public class CreateList
{
//Represent a node of the singly linked list
public class Node<T>{
public T data;
public Node<T> next;

public Node(T value) {
data = value;
next = null;
}
}

public class Palindrome<T>{
public Node<T> tail = null;
public int size;

//Create a new node
Node<T> newNode = new Node<T>(data);

//Checks if the list is empty
//If list is empty, both head and tail will point to new node
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
public Node<T> reverseList(Node<T> temp){
Node<T> current = temp;
Node<T> 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.
public void isPalindrome(){
Boolean 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

//Compare nodes of first half and second half of list
flag = false;
break;
}
}

if(flag)
Console.WriteLine("Given singly linked list is a palindrome");
else
Console.WriteLine("Given singly linked list is not a palindrome");
}

//display() will display all the nodes present in the list
public void display() {
//Node current will point to head

Console.WriteLine("List is empty");
return;
}
Console.WriteLine("Nodes of singly linked list: ");
while(current != null) {
//Prints each node by incrementing pointer
Console.Write(current.data + " ");
current = current.next;
}
Console.WriteLine();
}
}

public static void Main()
{
Palindrome<int> sList = new Palindrome<int>();

sList.display();

//Checks whether given list is palindrome or not
sList.isPalindrome();
}
}
```

Output:

```Nodes of singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
```

PHP

```<!DOCTYPE html>
<html>
<body>
<?php
//Represent a node of singly linked list
class Node{
public \$data;
public \$next;

function __construct(\$data){
\$this->data = \$data;
\$this->next = NULL;
}
}
class Palindrome{
public \$tail;
function __construct(){
\$this->tail = NULL;
\$this->size = 0;
}

//Create a new node
\$newNode = new Node(\$data);

//Checks if the list is empty
//If list is empty, both head and tail will point to new node
\$this->tail = \$newNode;
}
else {
//newNode will be added after tail such that tail's next will point to newNode
\$this->tail->next = \$newNode;
//newNode will become new tail of the list
\$this->tail = \$newNode;
}
//Size will count the number of nodes present in the list
\$this->size++;
}

//reverseList() will reverse the singly linked list and return the head of the list
function reverseList(\$temp){
\$current = \$temp;
\$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.
function isPalindrome(){
\$flag = true;

//Store the mid position of the list
\$mid = (\$this->size%2 == 0)? (\$this->size/2) : ((\$this->size+1)/2);

//Finds the middle node in given singly linked list
for(\$i=1; \$i<\$mid; \$i++){
\$current = \$current->next;
}

//Reverse the list after middle node to end

//Compare nodes of first half and second half of list
\$flag = false;
break;
}
}

if(\$flag)
print("Given singly linked list is a palindrome <br>");
else
print("Given singly linked list is not a palindrome <br>");
}

//display() will display all the nodes present in the list
function display() {
//Node current will point to head

print("List is empty <br>");
return;
}
print("Nodes of singly linked list: <br>");
while(\$current != NULL) {
//Prints each node by incrementing pointer
print(\$current->data . " ");
\$current = \$current->next;
}
print("<br>");
}
}

\$sList = new Palindrome();

\$sList->display();

//Checks whether given list is palindrome or not
\$sList->isPalindrome();
?>
</body>
</html>
```

Output:

``` Nodes of singly linked list:
1 2 3 2 1
Given singly linked list is a palindrome
```

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