Singly Linked List Program in java ~ Multiplex Coder

Question:

Bob and Alice are friends and interested in computer programming. One day Alice challenges Bob to create a singly linked list of N integers and find out the list is a palindrome or not. Return 1 or 0 denoting if it is a palindrome or not, respectively. Help Bob to win the challenge.

Note: Create the main() inside the class ‘ListDriver‘

Input format:
The first line of input contains an integer N denoting the size of the linked list. In the next line are N space separated values of the nodes of the linked list.

Output format:
For each test case output will be 1 if the linked list is a palindrome else 0.

Sample Input 1:
4
1 2 2 1

Sample Output1:

Sample Input 2:
9
3 4 2 5 6 2 7 1 9

Sample Output 2:

CODE:–

ListDriver.java

import java.util.Scanner;
public class ListDriver
{ class Node { int data;
Node next;
public Node(int data)
{ this.data=data;
this.next=null;
}
}
public int size;
public Node head= null;
public Node tail=null;
public void addNode(int data)
{ Node newNode=new Node(data);
if(head==null)
{ head = newNode;
tail=newNode;
}
else{tail.next=newNode;
tail =newNode;
}
size++;
}
public Node reverseList(Node temp)
{ Node current=temp;
Node prevNode=null, nextNode=null;
while(current!=null)
{nextNode=current.next;
current.next=prevNode;
prevNode=current;
current=nextNode;
}
return prevNode;
}
public void isPalindrome()
{ Node current= head;
boolean flag=true;
int mid=(size%2==0)?(size/2):((size+1)/2);
for(int i=1;i<mid;i++)
{current=current.next;
}
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)
System.out.println("1");
else
System.out.println("0");
}
public void display()
{ Node current=head;
if(head==null)
{ return;
}
while(current!=null)
{
current=current.next;
}
System.out.println();
}

public static void main(String[] args)
{ ListDriver sList=new ListDriver();
Scanner s= new Scanner(System.in);
int n;
n=s.nextInt();
do{sList.addNode(s.nextInt());
n--;
} while(n>=1);

sList.display();
sList.isPalindrome();
}
}