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.

Program to determine whether a singly linked list is the palindrome

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

Create a class Node which has two attributes: data and next. Next is a pointer to the next node in the list.
Create another class Palindrome which has three attributes: head, tail, and size.
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.
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.
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.
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.

Solution

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 palindromen”);      else          printf(“Given singly linked list is not a palindromen”);  }        //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 emptyn”);          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

Next Topic#

Program To Determine Whether A Singly Linked List Is The Palindrome