Having a bit of trouble adding a node to the end of my linked list. It only seems to display the very last one I added before I call my addFirst method. To me it looks like on the addLast method I'm trying to first create the node to assign it 5, then for the following numbers use a while loop to assign them to the last node on the linked list. Little stuck on why I can't get my output to display 5 and 6.
class LinkedList
{
private class Node
{
private Node link;
private int x;
}
//----------------------------------
private Node first = null;
//----------------------------------
public void addFirst(int d)
{
Node newNode = new Node();
newNode.x = d;
newNode.link = first;
first = newNode;
}
//----------------------------------
public void addLast(int d)
{
first = new Node();
if (first == null)
{
first = first.link;
}
Node newLast = new Node();
while (first.link != null)
{
first = first.link;
}
newLast.x = d;
first.link = newLast;
first = newLast;
}
//----------------------------------
public void traverse()
{
Node p = first;
while (p != null)
{
System.out.println(p.x);
p = p.link;
}
}
}
//==============================================
class test123
{
public static void main(String[] args)
{
LinkedList list = new LinkedList();
list.addLast(5);
list.addLast(6);
list.addLast(7);
list.addFirst(1);
list.addFirst(2);
list.addFirst(3);
System.out.println("Numbers on list");
list.traverse();
}
}
I've also tried creating a last Node and in the traverse method using a separate loop to traverse the last node. I end up with the same output!
public void addLast(int d)
{
Node newLast = new Node();
while (last.link != null)
{
last = newLast.link;
}
newLast.x = d;
newLast.link = last;
last = newLast;
}
The logic of your addLast method was wrong. Your method was reassigning first with every call the logic falls apart from that point forward. This method will create the Node for last and if the list is empty simply assign first to the new node last. If first is not null it will traverse the list until it finds a Node with a null link and make the assignment for that Nodes link.
public void addLast(int d) {
Node last = new Node();
last.x = d;
Node node = first;
if (first == null) {
first = last;
} else {
while (node.link != null) {
node = node.link;
}
node.link = last;
}
}
Your addLast() method displays the value of the last node because every time you append a node to the end of your list, you are overwriting the reference to "first". You are also doing this when you assign a new reference to first in the following line:
first = new Node();
Try the following:
public void addLast(int d)
{
Node newLast = new Node();
if (first == null)
{
newLast.x = d;
first = newLast;
return;
}
Node curr = first;
while (curr.link != null)
{
curr = curr.link;
}
newLast.x = d;
curr.link = newLast;
}
The method creates a new node and it is added to the end of the list after checking two conditions:
1.) If first is null: in this case the list is empty and the first node should be
initialized to the new node you are creating, then it will return (or you could do
an if-else).
2.) If first is not null: If the list is not empty you'll loop through your list until you
end up with a reference to the last node, after which you will set its next node to the
one you just created.
If you wanted to keep track of the tail of your list called "last", like you mentioned above, then add:
last = newLast
at the end of your method. Hope this helps!
Related
I'm studying for an exam, and this is a problem from an old test:
We have a singly linked list with a list head with the following declaration:
class Node {
Object data;
Node next;
Node(Object d,Node n) {
data = d;
next = n;
}
}
Write a method void addLast(Node header, Object x) that adds x at the end of the list.
I know that if I actually had something like:
LinkedList someList = new LinkedList();
I could just add items to the end by doing:
list.addLast(x);
But how can I do it here?
class Node {
Object data;
Node next;
Node(Object d,Node n) {
data = d ;
next = n ;
}
public static Node addLast(Node header, Object x) {
// save the reference to the header so we can return it.
Node ret = header;
// check base case, header is null.
if (header == null) {
return new Node(x, null);
}
// loop until we find the end of the list
while ((header.next != null)) {
header = header.next;
}
// set the new node to the Object x, next will be null.
header.next = new Node(x, null);
return ret;
}
}
You want to navigate through the entire linked list using a loop and checking the "next" value for each node. The last node will be the one whose next value is null. Simply make this node's next value a new node which you create with the input data.
node temp = first; // starts with the first node.
while (temp.next != null)
{
temp = temp.next;
}
temp.next = new Node(header, x);
That's the basic idea. This is of course, pseudo code, but it should be simple enough to implement.
public static Node insertNodeAtTail(Node head,Object data) {
Node node = new Node(data);
node.next = null;
if (head == null){
return node;
}
else{
Node temp = head;
while(temp.next != null){
temp = temp.next;
}
temp.next = node;
return head;
}
}
If you keep track of the tail node, you don't need to loop through every element in the list.
Just update the tail to point to the new node:
AddValueToListEnd(value) {
var node = new Node(value);
if(!this.head) { //if the list is empty, set head and tail to this first node
this.head = node;
this.tail = node;
} else {
this.tail.next = node; //point old tail to new node
}
this.tail = node; //now set the new node as the new tail
}
In plain English:
Create a new node with the given value
If the list is empty, point head and tail to the new node
If the list is not empty, set the old tail.next to be the new node
In either case, update the tail pointer to be the new node
Here is a partial solution to your linked list class, I have left the rest of the implementation to you, and also left the good suggestion to add a tail node as part of the linked list to you as well.
The node file :
public class Node
{
private Object data;
private Node next;
public Node(Object d)
{
data = d ;
next = null;
}
public Object GetItem()
{
return data;
}
public Node GetNext()
{
return next;
}
public void SetNext(Node toAppend)
{
next = toAppend;
}
}
And here is a Linked List file :
public class LL
{
private Node head;
public LL()
{
head = null;
}
public void AddToEnd(String x)
{
Node current = head;
// as you mentioned, this is the base case
if(current == null) {
head = new Node(x);
head.SetNext(null);
}
// you should understand this part thoroughly :
// this is the code that traverses the list.
// the germane thing to see is that when the
// link to the next node is null, we are at the
// end of the list.
else {
while(current.GetNext() != null)
current = current.GetNext();
// add new node at the end
Node toAppend = new Node(x);
current.SetNext(toAppend);
}
}
}
loop to the last element of the linked list which have next pointer to null then modify the next pointer to point to a new node which has the data=object and next pointer = null
Here's a hint, you have a graph of nodes in the linked list, and you always keep a reference to head which is the first node in the linkedList.
next points to the next node in the linkedlist, so when next is null you are at the end of the list.
The addLast() needs some optimisation as the while loop inside addLast() has O(n) complexity. Below is my implementation of LinkedList. Run the code with ll.addLastx(i) once and run it with ll.addLast(i) again , you can see their is a lot of difference in processing time of addLastx() with addLast().
Node.java
package in.datastructure.java.LinkedList;
/**
* Created by abhishek.panda on 07/07/17.
*/
public final class Node {
int data;
Node next;
Node (int data){
this.data = data;
}
public String toString(){
return this.data+"--"+ this.next;
}
}
LinkedList.java
package in.datastructure.java.LinkedList;
import java.util.ArrayList;
import java.util.Date;
public class LinkedList {
Node head;
Node lastx;
/**
* #description To append node at end looping all nodes from head
* #param data
*/
public void addLast(int data){
if(head == null){
head = new Node(data);
return;
}
Node last = head;
while(last.next != null) {
last = last.next;
}
last.next = new Node(data);
}
/**
* #description This keep track on last node and append to it
* #param data
*/
public void addLastx(int data){
if(head == null){
head = new Node(data);
lastx = head;
return;
}
if(lastx.next == null){
lastx.next = new Node(data);
lastx = lastx.next;
}
}
public String toString(){
ArrayList<Integer> arrayList = new ArrayList<Integer>(10);
Node current = head;
while(current.next != null) {
arrayList.add(current.data);
current = current.next;
}
if(current.next == null) {
arrayList.add(current.data);
}
return arrayList.toString();
}
public static void main(String[] args) {
LinkedList ll = new LinkedList();
/**
* #description Checking the code optimization of append code
*/
Date startTime = new Date();
for (int i = 0 ; i < 100000 ; i++){
ll.addLastx(i);
}
Date endTime = new Date();
System.out.println("To total processing time : " + (endTime.getTime()-startTime.getTime()));
System.out.println(ll.toString());
}
}
The above programs might give you NullPointerException. This is an easier way to add an element to the end of linkedList.
public class LinkedList {
Node head;
public static class Node{
int data;
Node next;
Node(int item){
data = item;
next = null;
}
}
public static void main(String args[]){
LinkedList ll = new LinkedList();
ll.head = new Node(1);
Node second = new Node(2);
Node third = new Node(3);
Node fourth = new Node(4);
ll.head.next = second;
second.next = third;
third.next = fourth;
fourth.next = null;
ll.printList();
System.out.println("Add element 100 to the last");
ll.addLast(100);
ll.printList();
}
public void printList(){
Node t = head;
while(n != null){
System.out.println(t.data);
t = t.next;
}
}
public void addLast(int item){
Node new_item = new Node(item);
if(head == null){
head = new_item;
return;
}
new_item.next = null;
Node last = head;
Node temp = null;
while(last != null){
if(last != null)
temp = last;
last = last.next;
}
temp.next = new_item;
return;
}
}
I wrote a method that adds an item to the front of a doubly linked list. Every time I call the function, it should add the item passed through it. The function works as expected, except for when I iterate through the doubly linked list and print each item, it always prints null twice at the end. My code is below.
public Deque() {
first = new Node();
last = new Node();
first.next = last;
last.prev = first;
}
public void addFirst(E item) {
if (item.equals(null)) {
throw new NullPointerException();
} else {
if (first.equals(null) && last.equals(null)) {
first = new Node();
first.next.item = item;
first.next.next = null;
last = first;
} else {
Node node = new Node();
node.item = item;
node.next = first;
first = node;
}
}
N++;
}
public static void main(String[] args) {
Deque<Integer> lst = new Deque<Integer>(); // empty list
lst.addFirst(1);
lst.addFirst(5);
lst.addFirst(7);
lst.addFirst(9);
Iterator<Integer> it = lst.iterator(); // tests iterator method
while (it.hasNext()) {
Integer val = it.next();
System.out.println(val);
}
}
That code prints: 9, 5, 7, 1, null, null. I cannot figure out the two extra null values are being printed. Can anybody tell me how I can fix my code so that it does not print null two times at the end?
You're creating two nodes called first and last as part of the constructor of the Deque class.
public Deque() {
first = new Node();
last = new Node();
first.next = last;
last.prev = first;
}
So, your items get added in front of them. When you print them out, it prints the ones you added and then first and last, and since they don't have an item set, it prints null.
I'm working on an assignment for my Data Structures class. We have to create an address book using our own sorted linked based list adt. Right now the add method works, but it seems to make all the nodes point to the first node. Whenever I try to output the the list using getEntry() in a for loop, it gives me the last added entry each time. I've tried using toArray but it does the same thing. Can you see any problems?
public class GTSortedLinkedBasedList implements GTListADTInterface {
private Node firstNode;
private int numberOfEntries;
public GTSortedLinkedBasedList(){
//firstNode = new Node(null);
numberOfEntries = 0;
}
public void setNumberOfEntries(int x){
numberOfEntries = x;
}
public void add(ExtPersonType newEntry){
//firstNode = null;
Node newNode = new Node(newEntry);
Node nodeBefore = getNodeBefore(newEntry);
if (isEmpty() || (nodeBefore == null))
{
// Add at beginning
newNode.setNextNode(firstNode);
firstNode = newNode;
}
else
{
// Add after nodeBefore
Node nodeAfter = nodeBefore.getNextNode();
newNode.setNextNode(nodeAfter);
nodeBefore.setNextNode(newNode);
} // end if
numberOfEntries++;
}
private Node getNodeBefore(ExtPersonType anEntry){
Node currentNode = getFirstNode();
Node nodeBefore = null;
while ((currentNode != null) &&
(anEntry.getFirstName().compareTo(currentNode.getData().getFirstName()) > 0))
{
nodeBefore = currentNode;
currentNode = currentNode.getNextNode();
} // end while
return nodeBefore;
}
private class Node {
private ExtPersonType data;
private Node next;
public Node(ExtPersonType dataValue) {
next = null;
data = dataValue;
}
public Node(ExtPersonType dataValue, Node nextValue) {
next = nextValue;
data = dataValue;
}
public ExtPersonType getData(){
return data;
}
public void setData(ExtPersonType newData){
data = newData;
}
public Node getNextNode(){
return next;
}
public void setNextNode(Node newNode){
next = newNode;
}
}
public ExtPersonType getEntry(int givenPosition) {
if ((givenPosition >= 1) && (givenPosition <= numberOfEntries)){
assert !isEmpty();
return getNodeAt(givenPosition).getData();
}
else{
throw new IndexOutOfBoundsException("Illegal position given to getEntry operation.");
}
}
public void loadData(GTSortedLinkedBasedList contacts) throws FileNotFoundException{
//int index = 0;
ExtPersonType person = new ExtPersonType();
DateType tempDate = new DateType();
AddressType tempAddress = new AddressType();
Scanner file = new Scanner(new FileInputStream("Programming Assignment 1 Data.txt"));
while(file.hasNext()){
person.setFirstName(file.next());
person.setLastName(file.next());
tempDate.setMonth(file.nextInt());
tempDate.setDay(file.nextInt());
tempDate.setYear(file.nextInt());
person.setDOB(tempDate);
tempAddress.setStreetAddress(file.nextLine());
if(tempAddress.getStreetAddress().isEmpty()){
tempAddress.setStreetAddress(file.nextLine());
}
tempAddress.setCity(file.nextLine());
tempAddress.setState(file.nextLine());
tempAddress.setZipCode(file.nextLine());
person.setAddress(tempAddress);
person.setPhoneNumber(file.nextLine());
person.setPersonStatus(file.nextLine());
if(person.getPersonStatus().isEmpty()){
person.setPersonStatus(file.nextLine());
}
contacts.add(person);
System.out.println(contacts.getEntry(contacts.getLength()).getFirstName());
//index++;
}
}
public static void main(String[] args) throws FileNotFoundException {
AddressBook ab = new AddressBook();
ab.loadData(ab);
ExtPersonType people = new ExtPersonType();
//people = ab.toArray(people);
System.out.println(ab.getLength());
for(int cnt = 1; cnt <= ab.getLength(); cnt++){
people = ab.getEntry(cnt);
System.out.println(people.getFirstName());
}
}
EDIT: The add method is overwriting each previous object with the newly added one. It also doesn't seem to matter if I do a sorted list or just a basic list.
I'm not going to lie here, I'm not totally sure I understand your code but I think I see what's wrong. In your getNodeBefore() method's code, you set currentNode() always to firstNode(). I believe that is causing the problem. I see that you are trying to recursively move through the list to find the proper node but I don't think each recursive call is causing movement through the list. I suggest you add properties to the object that represent the forward and backward nodes.
Something like this...
private T data;
private Node nodeBefore;
private Node nodeAfter;
As you create objects, you assign the properties before and after and then all the information you need is contained in the object itself.
To move recursively through the list you would then just add a statement like currentNode = currentNode.nodeAfter.
Your getNodeBefore() method would simply return currentNode.nodeBefore and getNodeAfter() would return currentNode.nodeAfter.
You don't have code that handles the situation where the node being added will be the first node in the list, but the list is also not empty. In this case, getNodeBefore returns null, and your code overwrites the root node.
Try
if (isEmpty() && (nodeBefore == null))
{
// Add at beginning
newNode.setNextNode(firstNode);
firstNode = newNode;
}
else if(nodeBefore == null)
{
Node temp = new Node();
temp.setNextNode(first.next);
temp.setData(first.data);
newNode.setNextNode(temp);
firstNode = newNode;
}
I have a question for combining two linkedlist. Basically, I want to append one linkedlist to the other linkedlist.
Here is my solution. Is there a more efficient way to do it without looping the first linkedlist? Any suggestion would be appreciated.
static Node connect(LinkedList list1, LinkedList list2) {
Node original = list1.first;
Node previous = null;
Node current = list1.first;
while (current != null) {
previous = current;
current = current.next;
}
previous.next = list2.first;
return original;
}
Use list1.addAll(list2) to append list2 at the end of list1.
For linked lists, linkedList.addAll(otherlist) seems to be a very poor choice.
the java api version of linkedList.addAll begins:
public boolean addAll(int index, Collection<? extends E> c) {
checkPositionIndex(index);
Object[] a = c.toArray();
so even when you have 2 linked lists, the second one gets converted to an array, then re-constituted into individual elements. This is worse than just merging 2 arrays.
I guess this is your own linked list implementation? With only a pointer to next element, the only way to append at the end is to walk all the elements of the first list.
However, you could store a pointer to the last element to make this operation run in constant time (just remember to update the last element of the new list to be the last element of the added list).
The best way is to append the second list to the first list.
1. Create a Node Class.
2. Create New LinkedList Class.
public class LinkedList<T> {
public Node<T> head = null;
public LinkedList() {}
public void addNode(T data){
if(head == null) {
head = new Node<T>(data);
} else {
Node<T> curr = head;
while(curr.getNext() != null) {
curr = curr.getNext();
}
curr.setNext(new Node<T>(data));
}
}
public void appendList(LinkedList<T> linkedList) {
if(linkedList.head == null) {
return;
} else {
Node<T> curr = linkedList.head;
while(curr != null) {
addNode((T) curr.getData());
curr = curr.getNext();
}
}
}
}
3. In the Main function or whereever you want this append to happen, do it like this.
LinkedList<Integer> n = new LinkedListNode().new LinkedList<Integer>();
n.addNode(23);
n.addNode(41);
LinkedList<Integer> n1 = new LinkedListNode().new LinkedList<Integer>();
n1.addNode(50);
n1.addNode(34);
n.appendList(n1);
I like doing this way so that there isn't any need for you to pass both these and loop again in the first LinkedList.
Hope that helps
My Total Code:
NOTE: WITHOUT USING JAVA API
class Node {
Node next;
int data;
Node(int d){
data = d;
next = null;
}
}
public class OddEvenList {
Node head;
public void push(int new_data){
Node new_node = new Node(new_data);
new_node.next = head;
head = new_node;
}
Node reverse(Node head){
Node prev = null;
Node next = null;
Node curr = head;
while(curr != null){
next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
head = prev;
return head;
}
Node merge(Node head1, Node head2){
Node curr_odd = head1;
Node curr_even = head2;
Node prev = null;
while(curr_odd != null){
prev = curr_odd;
curr_odd = curr_odd.next;
}
prev.next = curr_even;
return head1;
}
public void print(Node head){
Node tnode = head;
while(tnode != null){
System.out.print(tnode.data + " -> ");
tnode = tnode.next;
}
System.out.println("Null");
}
public static void main(String[] args) {
// TODO Auto-generated method stub
OddEvenList odd = new OddEvenList();
OddEvenList even = new OddEvenList();
OddEvenList merge = new OddEvenList();
odd.push(1);
odd.push(3);
odd.push(5);
odd.push(7);
odd.push(9);
System.out.println("Odd List: ");
odd.print(odd.head);
System.out.println("Even List: ");
even.push(0);
even.push(2);
even.push(4);
even.push(6);
even.push(8);
even.print(even.head);
System.out.println("After Revrse: --------------------");
Node node_odd =odd.reverse(odd.head);
Node node_even = even.reverse(even.head);
System.out.println("Odd List: ");
odd.print(node_odd);
System.out.println("Even List: ");
even.print(node_even);
System.out.println("Meged: --------------");
Node merged = merge.merge(node_odd, node_even);
merge.print(merged);
}
}
I'm studying for an exam, and this is a problem from an old test:
We have a singly linked list with a list head with the following declaration:
class Node {
Object data;
Node next;
Node(Object d,Node n) {
data = d;
next = n;
}
}
Write a method void addLast(Node header, Object x) that adds x at the end of the list.
I know that if I actually had something like:
LinkedList someList = new LinkedList();
I could just add items to the end by doing:
list.addLast(x);
But how can I do it here?
class Node {
Object data;
Node next;
Node(Object d,Node n) {
data = d ;
next = n ;
}
public static Node addLast(Node header, Object x) {
// save the reference to the header so we can return it.
Node ret = header;
// check base case, header is null.
if (header == null) {
return new Node(x, null);
}
// loop until we find the end of the list
while ((header.next != null)) {
header = header.next;
}
// set the new node to the Object x, next will be null.
header.next = new Node(x, null);
return ret;
}
}
You want to navigate through the entire linked list using a loop and checking the "next" value for each node. The last node will be the one whose next value is null. Simply make this node's next value a new node which you create with the input data.
node temp = first; // starts with the first node.
while (temp.next != null)
{
temp = temp.next;
}
temp.next = new Node(header, x);
That's the basic idea. This is of course, pseudo code, but it should be simple enough to implement.
public static Node insertNodeAtTail(Node head,Object data) {
Node node = new Node(data);
node.next = null;
if (head == null){
return node;
}
else{
Node temp = head;
while(temp.next != null){
temp = temp.next;
}
temp.next = node;
return head;
}
}
If you keep track of the tail node, you don't need to loop through every element in the list.
Just update the tail to point to the new node:
AddValueToListEnd(value) {
var node = new Node(value);
if(!this.head) { //if the list is empty, set head and tail to this first node
this.head = node;
this.tail = node;
} else {
this.tail.next = node; //point old tail to new node
}
this.tail = node; //now set the new node as the new tail
}
In plain English:
Create a new node with the given value
If the list is empty, point head and tail to the new node
If the list is not empty, set the old tail.next to be the new node
In either case, update the tail pointer to be the new node
Here is a partial solution to your linked list class, I have left the rest of the implementation to you, and also left the good suggestion to add a tail node as part of the linked list to you as well.
The node file :
public class Node
{
private Object data;
private Node next;
public Node(Object d)
{
data = d ;
next = null;
}
public Object GetItem()
{
return data;
}
public Node GetNext()
{
return next;
}
public void SetNext(Node toAppend)
{
next = toAppend;
}
}
And here is a Linked List file :
public class LL
{
private Node head;
public LL()
{
head = null;
}
public void AddToEnd(String x)
{
Node current = head;
// as you mentioned, this is the base case
if(current == null) {
head = new Node(x);
head.SetNext(null);
}
// you should understand this part thoroughly :
// this is the code that traverses the list.
// the germane thing to see is that when the
// link to the next node is null, we are at the
// end of the list.
else {
while(current.GetNext() != null)
current = current.GetNext();
// add new node at the end
Node toAppend = new Node(x);
current.SetNext(toAppend);
}
}
}
loop to the last element of the linked list which have next pointer to null then modify the next pointer to point to a new node which has the data=object and next pointer = null
Here's a hint, you have a graph of nodes in the linked list, and you always keep a reference to head which is the first node in the linkedList.
next points to the next node in the linkedlist, so when next is null you are at the end of the list.
The addLast() needs some optimisation as the while loop inside addLast() has O(n) complexity. Below is my implementation of LinkedList. Run the code with ll.addLastx(i) once and run it with ll.addLast(i) again , you can see their is a lot of difference in processing time of addLastx() with addLast().
Node.java
package in.datastructure.java.LinkedList;
/**
* Created by abhishek.panda on 07/07/17.
*/
public final class Node {
int data;
Node next;
Node (int data){
this.data = data;
}
public String toString(){
return this.data+"--"+ this.next;
}
}
LinkedList.java
package in.datastructure.java.LinkedList;
import java.util.ArrayList;
import java.util.Date;
public class LinkedList {
Node head;
Node lastx;
/**
* #description To append node at end looping all nodes from head
* #param data
*/
public void addLast(int data){
if(head == null){
head = new Node(data);
return;
}
Node last = head;
while(last.next != null) {
last = last.next;
}
last.next = new Node(data);
}
/**
* #description This keep track on last node and append to it
* #param data
*/
public void addLastx(int data){
if(head == null){
head = new Node(data);
lastx = head;
return;
}
if(lastx.next == null){
lastx.next = new Node(data);
lastx = lastx.next;
}
}
public String toString(){
ArrayList<Integer> arrayList = new ArrayList<Integer>(10);
Node current = head;
while(current.next != null) {
arrayList.add(current.data);
current = current.next;
}
if(current.next == null) {
arrayList.add(current.data);
}
return arrayList.toString();
}
public static void main(String[] args) {
LinkedList ll = new LinkedList();
/**
* #description Checking the code optimization of append code
*/
Date startTime = new Date();
for (int i = 0 ; i < 100000 ; i++){
ll.addLastx(i);
}
Date endTime = new Date();
System.out.println("To total processing time : " + (endTime.getTime()-startTime.getTime()));
System.out.println(ll.toString());
}
}
The above programs might give you NullPointerException. This is an easier way to add an element to the end of linkedList.
public class LinkedList {
Node head;
public static class Node{
int data;
Node next;
Node(int item){
data = item;
next = null;
}
}
public static void main(String args[]){
LinkedList ll = new LinkedList();
ll.head = new Node(1);
Node second = new Node(2);
Node third = new Node(3);
Node fourth = new Node(4);
ll.head.next = second;
second.next = third;
third.next = fourth;
fourth.next = null;
ll.printList();
System.out.println("Add element 100 to the last");
ll.addLast(100);
ll.printList();
}
public void printList(){
Node t = head;
while(n != null){
System.out.println(t.data);
t = t.next;
}
}
public void addLast(int item){
Node new_item = new Node(item);
if(head == null){
head = new_item;
return;
}
new_item.next = null;
Node last = head;
Node temp = null;
while(last != null){
if(last != null)
temp = last;
last = last.next;
}
temp.next = new_item;
return;
}
}