NullPointerException Error using linked lists

Question

I just finished up working on this program and got it to compile but it breaks after user input and gives me this:

Please input 0 or more values at keyboard 12 4 3 2 1

Exception in thread "main" java.lang.NullPointerException
at Search.buildList(Search.java:41)
at Search.main(Search.java:10)

Here is the code:

import java.io.*; 
import java.util.*;

public class Search { 
public static void main(String argv[]) throws IOException { 

Scanner stdin = new Scanner(System.in);
System.out.println("Please input 0 or more values at keyboard");
Node head = buildList();

System.out.println("Now printing list");
printList(head);
System.out.println("\nWhat key in list are you searching for? ");
int key = stdin.nextInt();
System.out.print("Your key was ");
if (search(head, key))
System.out.println("found.");
else
System.out.println("not found.");

}

private static void printList(Node head)
{
            if (head != null)
            {
                    System.out.print(head.getItem() + " ");
                    printList(head.getNext());
            }
}

private static Node buildList() throws IOException
{
 // Post : Inserts 0 or more numerical values from keyboard into list
//          using the Scanner class and returns head of list

Scanner input = new Scanner(System.in);
Node head = null;
Node first = new Node(input.nextInt());
head.setNext(first);
while(input.hasNext())
{    
insert(first, input.nextInt());
/*
  Node curr = new Node(input.nextInt());
  Node prev = head;
  while (true)
  {
prev = prev.getNext();
if ((int)curr.getItem() < (int)prev.getItem())
{
  head.setNext(curr);
  curr.setNext(prev);
  break;
}
if (prev.getNext() == null)
{
  prev.setNext(curr);
  break;
}
  }*/
}
return first;
} 

private static Node insert(Node head, Comparable newValue)
{
Node prev, curr = head;

for (prev = null,  curr = head;
         curr != null && newValue.compareTo(curr.getItem()) > 0;
         prev = curr, curr = curr.getNext() ) {}

    Node newNode = new Node(newValue, curr);
if (prev != null)
    {
        prev.setNext(newNode);
    return head;
    }
else
    return newNode;
}

private static boolean search(Node head, Comparable key)
{
 // PRE:  head points to the front of linked list;  list may be
 //         empty or non-empty;  key is item searching for
 // POST: returns true or false regarding whether key is found in
 //         list
if (head == null){
    return false;}
else if (head.getItem().equals(key)){
    return true;}
else {
    return search(head.getNext(), key);
}

} 

}

Any ideas?

The output should be similar to the following:

Please input 0 or more values at keyboard

12 4 -1 5 3 0 2

Now printing list

-1 0 2 3 4 5 12 What key are you searching for? 15 Your key was not found

Answer 1

Node head = null;

whenever you call a method on a null object you get an nullPointerException.That is why head.setNext(first); is giving you exception. so instead of this you can do

Node head = new Node();

you will avoid NullPointerException with this.

According to your requirement you should do this.

private static Node buildList() throws IOException
{
 // Post : Inserts 0 or more numerical values from keyboard into list
//          using the Scanner class and returns head of list

Scanner input = new Scanner(System.in);
Node head = null;
Node first = new Node(input.nextInt());
head=first; //assigning the first value to head
while(input.hasNext())
{    
insert(first, input.nextInt());
head.setNext(first);//insert the node in the list
}
return first;
} 

Note: I am assuming that setNext() inserts the node in the list at appropriate location not directly in the next position of head node (otherwise you will get only 2 nodes no matter how many numbers you insert)

Answer 2

Node head = null;

The above line will make the head which is an object reference variable of type Node to null, now calling any method on this object reference variable will lead to NullPointerException.

Node head = new Node();

The line will be a better approach as this will prevent the NullPointerException.