Q:

Program to determine whether a singly linked list is the palindrome

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robaa
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2022-11-10
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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.
  3. addNode() will add a new node to the list:
    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.

Input:

 

#Add nodes to the list  

sList.addNode(1);  

sList.addNode(2);  

sList.addNode(3);  

sList.addNode(2);   

sList.addNode(1);   

Output:

Nodes of the singly linked list: 1 2 3 2 1
Given singly linked list is a palindrome
Python programming, Php programming, Java programming, C programming, c# programming, examples, Python programming examples, Php programming examples, Java programming examples, C programming examples, c# programming examples,
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robaa
2022-11-10
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Python

#Represent a node of the singly linked list  
class Node:  
    def __init__(self,data):  
        self.data = data;  
        self.next = None;  
          
class Palindrome:  
    #Represent the head and tail of the singly linked list  
    def __init__(self):  
        self.head = None;  
        self.tail = None;  
        self.size = 0;  
          
    #addNode() will add a new node to the list  
    def addNode(self, data):  
        #Create a new node  
        newNode = Node(data);  
          
        #Checks if the list is empty  
        if(self.head == None):  
            #If list is empty, both head and tail will point to new node  
            self.head = newNode;  
            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):  
        current = self.head;  
        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  
        revHead = self.reverseList(current.next);  
          
        #Compare nodes of first half and second half of list  
        while(self.head != None and revHead != None):  
            if(self.head.data != revHead.data):  
                flag = False;  
                break;  
                  
            self.head = self.head.next;  
            revHead = revHead.next;  
              
        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  
        current = self.head;  
          
        if(self.head == None):  
            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();  
   
#Add nodes to the list  
sList.addNode(1);  
sList.addNode(2);  
sList.addNode(3);  
sList.addNode(2);  
sList.addNode(1);  
   
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;  
};      
   
//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(1);  
      
    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;  
    //Represent the head and tail of the singly linked list  
    public Node head = null;  
    public Node tail = null;  
      
    //addNode() will add a new node to the list  
    public void addNode(int data) {  
        //Create a new node  
        Node newNode = new Node(data);  
          
        //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  
    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(){  
        Node current = head;  
        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  
        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)  
            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  
            Node current = head;  
              
            if(head == null) {  
                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();  
          
        //Add nodes to the list  
        sList.addNode(1);  
        sList.addNode(2);  
        sList.addNode(3);  
        sList.addNode(2);  
        sList.addNode(1);  
          
        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>{  
        //Represent the head and tail of the singly linked list  
        public Node<T> head = null;               
         public Node<T> tail = null;  
        public int size;  
          
        //addNode() will add a new node to the list  
        public void addNode(T data) {  
            //Create a new node  
            Node<T> newNode = new Node<T>(data);  
   
            //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  
        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(){  
            Node<T> current=head;  
            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  
            Node<T> revHead = reverseList(current.next);  
   
            //Compare nodes of first half and second half of list  
            while(head != null && revHead != null){  
                if(!(head.data.Equals(revHead.data))){  
                    flag = false;  
                    break;  
                }  
                head = head.next;  
                revHead = revHead.next;  
            }  
   
            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  
            Node<T> current = head;  
              
            if(head == null) {  
                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>();  
          
        //Add nodes to the list  
        sList.addNode(1);  
        sList.addNode(2);  
        sList.addNode(3);  
        sList.addNode(2);  
        sList.addNode(1);  
          
        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{  
    //Represent the head and tail of the singly linked list  
    public $head;  
    public $tail;  
    function __construct(){  
        $this->head = NULL;  
        $this->tail = NULL;  
        $this->size = 0;  
    }  
      
    //addNode() will add a new node to the list  
    function addNode($data) {  
        //Create a new node  
        $newNode = new Node($data);  
          
        //Checks if the list is empty  
        if($this->head == NULL) {  
            //If list is empty, both head and tail will point to new node  
            $this->head = $newNode;  
            $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(){  
        $current = $this->head;  
        $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  
        $revHead = $this->reverseList($current->next);  
   
        //Compare nodes of first half and second half of list  
        while($this->head != null && $revHead != null){  
            if($this->head->data != $revHead->data){  
                $flag = false;  
                break;  
            }  
            $this->head = $this->head->next;  
            $revHead = $revHead->next;  
        }  
   
        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  
        $current = $this->head;  
          
        if($this->head == NULL) {  
            print("List is empty <br>");  
            return;  
        }  
        print("Nodes of singly linked list: <br>");  
        while($current != NULL) {  
            //Prints each node by incrementing pointer  
            print($current->data . " ");  
            $current = $current->next;  
        }  
        print("<br>");  
    }  
}  
      
$sList = new Palindrome();  
          
//Add nodes to the list  
$sList->addNode(1);  
$sList->addNode(2);  
$sList->addNode(3);  
$sList->addNode(2);  
$sList->addNode(1);  
   
$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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Singly Linked List Programs
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