I am trying to print a linked list. Both head and node are references of each other but still, the head is not changing after while loop in printLL() function. After end of while loop node is null but head is still pointing to 10. Why this is happening?
class Node {
int data;
Node next;
Node(int value) {
this.data = value;
this.next = null;
}
}
class LL {
Node head;
void createLL() {
head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
}
void printLL() {
Node node = head;
System.out.println("Printing full Linked List");
while (node != null) {
System.out.println(node.data);
node = node.next;
}
System.out.println("Value of head is " + head.data);
if (head.equals(node)) {
System.out.println("head and node both are same");
}
else {
System.out.println("head and node are not same");`
}
}
}
// Output -
// Printing full Linked List
// 10
// 20
// 30
// 40
// Value of head is 10
// head and node are not same
What is a Linked List ?
A linked list is formed by nodes that are linked together like a chain. Each node holds data, along with a pointer to the next node in the list.
The following image shows the basic structure of a Singly Linked List.
As you can see the Singly Linked List contains a head node: a pointer pointing to the first element of the list. Whenever we want to traverse the list, we can do so by using this head node.
Now using the below method of createLL(), you have created a linked list,
void createLL() {
head = new Node(10);
head.next = new Node(20);
head.next.next = new Node(30);
head.next.next.next = new Node(40);
}
This can be represented using an image as below for easier understanding,
So in the createLL method you have now created a linked list like
10->20->30->40->null
here the first node which is also called as the "head node" contains the data 10.
Now inside the method printLL(), the first line is as below,
Node node = head;
Here you created a temporary node named "node" and assigned it to value of head. The head contains the information of the next node and the next node contains the information of its next node and so on. So now using your temporary node "node" and assigning it the value of headnode, you can traverse the whole linked list you created via the method createLL().
And in the while loop below,
while (node != null) {
System.out.println(node.data);
node = node.next;
}
You have now traversed the linked list one by one until it reached to null. Remember the last node of a linked list always points to null. This while loop did not reassign the value/data of any nodes. But it just iterated from the first node till it reached the lastnode using a temporary node called "node". And the while loop finally assigns the "node" variable to null.
So in the if loop comparison,
if (head.equals(node))
you are comparing,
if(10.equals(null)
which will return as false, since headnode still pointing to the 10 itself. only the temporary node "node" is pointing to the null which does not affect the headnode or any node of the linked list you have created.
Related
You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.
Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.
class Solution {
/*Global variable pre to track the last node we visited */
Node pre = null;
public Node flatten(Node head) {
if (head == null) {
return null;
}
/*Connect last visted node with current node */
if (pre != null) {
pre.next = head;
head.prev = pre;
}
pre = head;
/*Store head.next in a next pointer in case recursive call to flatten head.child overrides head.next*/
Node next = head.next;
flatten(head.child);
head.child = null;
flatten(next);
return head;
}
}
For the question, the above leetcode solution works. But I don't understand this:
/*Store head.next in a next pointer in case recursive call to flatten head.child overrides head.next*/
Node next = head.next;
Can anyone explain this part? How does the head.child override head.next?
flatten(head.child) may change head.next due to the part where we connect last visited node with the current node:
if (pre != null) {
pre.next = head;
head.prev = pre;
}
At this stage, pre represents older head and head represents head.child. So we actually connect our head with its child. But if we do so, we will lose our connection to the actual head.next, therefore we have to keep it in an extra variable called next. That's why we keep it before calling the flatten() function.
I just started to learn the concept of Linked List recently, and have a problem on how the else statement works in the below code. How does it traverse to next node until a NULL value is found in the node?
public class LinkedList {
Node head;
public void insert(int data) {
Node node = new Node(data);
node.data = data;
if(head == null) {
head = node;
}else{
Node temp = head;
while(temp.next != null) {
temp = temp.next;
}
temp.next = node;
}
}
}
The Node class is the representation of a node in a linked list. Each Node has its own data and a pointer/reference to the next Node, which is stored in the next member of the Node class.
Now answering your question about traversal. We start the traversal from the head of the linked list - Node temp = head;. We now use a while loop to traverse the linked list and stop only when we reach the last Node in the linked list. The last node will have the next reference as null, hence the condition while(temp.next != null). Inside the loop, you are hopping to the next node and the loop will stop when you reach the last node. You then set the new Node as the next node of the last node in order to fulfill the data insertion function.
So, I have been studied about linked list and had created this insert method.
private void insert(Node head, int data)
{
Node node = new Node(data);
Node first = head;
if(first == null)
{
head = node;
}
else
{
node.nextLink = first;
head = node;
//System.out.println(node.data);
}
}
and this traverse method
public void traversingLinkedList(Node head)
{
Node current = head;
while(current != null)
{
int data = current.data;
System.out.println(data);
current = current.nextLink;
}
}
But it is not showing the node when I am inserting it.
The node data shows when I uncomment the print line in method insert.
for example,
LinkedList present is 10 -> 20 -> 30
after using insert(head,4)
I still get 10 -> 20 -> 30
though in the method insert when I uncheck the print method
it is showing first node data as 4
but when traversing it is not showing!
Why?
When calling a method in Java, the variables are copied, not referenced. This means that in your case, the variable head inside the insert method is only local and its modifications will not be visible outside the method.
Thus, since you are inserting elements at the front, the new head after the insertion is the node you have created (not the previous one) and you need to return it to update the next calls. Moreover, you could simplify the code of your insert method since you will always update the head value and the only conditional part is if there are more elements in the list or not.
private Node insert(Node head, int data)
{
Node node = new Node(data);
if (head != null) {
node.nextLink = head;
}
head = node;
return head;
}
In this case your main method should look like:
// LinkedList 10->20->30
head = new Node(30);
head = insert(head, 20);
head = insert(head, 10);
// Add the new 4 head: 4->10->20->30
head = insert(head, 4);
// Traversing
traversingLinkedList(head);
head is a local variable, so assigning values to it inside your insert(Node head, int data) doesn't affect the Node passed to the method.
If your insert method is part of some LinkedList class, that class should hold a reference to the head of the list, and insert should assign to that reference. In that case you won't need to pass Node head as argument to insert.
The only way you can modify the list by using the passed Node argument is if the method would change the nextLink of that Node.
emphasized text i need to pass header of the LINKED LIST and data to be inserted in that linked list.assuming the list is in sorted order i need to check data from each node and insert new node to give newly sorted list.
im getting null pointer exception ,, i need to know what im doing wrong
/*
Insert Node at the end of a linked list
head pointer input could be NULL as well for empty list
Node is defined as
class Node {
int data;
Node next;
Node prev;
}
*/
Node SortedInsert(Node head,int data) {
Node root= head;
if(head==null){
root.data=data;
root.next=null;
root.prev=null;
}else if(head.data>data){
Node newnode = new Node();
newnode.data=data;
newnode.next=head;
newnode.prev=null;
head.prev=newnode;
root=newnode;
}
int k=0;
while(head!=null && k==0){
if(head.data<data && head.next.data>data && head.next!=null){
Node temp=head.next;
Node newnode = new Node();
newnode.data=data;
newnode.next=temp;
newnode.prev=head;
head.next=newnode;
temp.prev=newnode;k++; break;
}
else if(head.data<data && head.next==null){
//Node temp=head.next;
Node newnode = new Node();
newnode.data=data;
newnode.next=null;
newnode.prev=head;
head.next=newnode;k++;break;
//temp.prev=newnode;
}else
{head=head.next;}
}
return root;
}
im getting null pointer exception at second if statement inside while loop.
I found some errors in your code which might be giving NullPointerException So change it accordingly.
First mistake is here:
Node root= head;
if(head==null){
root.data=data;
root.next=null;
root.prev=null;
}
So here you need to first create an object of Node class and assign it to root so code will look like :
Node root= head;
if(head==null){
root=new Node();
root.data=data;
root.next=null;
root.prev=null;
}
Another Mistake I encountered is in condition of if(head.data<data && head.next.data>data && head.next!=null).
Here you are should validate head.next before accessing it in head.next.data.
Suppose if head.next is null then the evaluation of condition of loop goes like this.
1) head.data<data so suppose this return true so we will check next condition.
2) head.next.data>data now if head.next is null then here condition would be null.data which will throw an NullPointerException. So here you should also check that head.next is not null. You are doing this is next condition but it is getting executed before validation it.
So here you just need to change order of the condition of if statement like: if(head.data<data && head.next!=null && head.next.data>data).
This will solve your problem.
Node root= head;
if(head==null){
root.data=data;
here you are trying to set data for a null object
you should allocate memory for root first
e.g.
head = new Node();
root = head
//then continue your code
I'm fairly new to JAVA and OOP and I'm currently following an academic course where I'm learning data structures and algorithms in java.
As I was learning about implementation of linked lists I've ran into a small problem of not understanding the code how to create a node when implementing a linked list(I'm familiar with constructors and bit of recursion ).
Code of the Node class as follows
public class Node {
public int info;
public Node next, prev;
public Node (int el) {
this (el,null,null);
}
public Node (int el,Node n,Node p){
info = el; next =n; prev=p;
}
}
I need to know what's happening behind the scene when the code executes(especially how the line3 works)and code of the List class is as follows
public class List {
private Node head, tail;
public List ( ){
head = tail = null;
}
public boolean isEmpty( ){
return head == null;
}
public void addToTail (int el) {
if (!isEmpty ( )) {
tail = new Node (el, null, tail);
tail.prev.next = tail;
}
else head = tail = new Node(el);
}
public int removeFromTail ( ){
int el = tail.info;
if (head == tail)
head = tail =null;
else
{
tail = tail.prev; tail.next = null;
}
return el;
}
}
(This example is given by the academy where I'm learning and I want to know how it works)
Please explain how the Node class works.
Ok let's start from the Node Class
public Node next, prev;
public Node (int el) {
this (el,null,null);
}
Here the objects next and prev are references to the next and previous nodes to the current node (which is your current object (this))
this (el,null,null);
It means you are creating a node which has no previous or next node. as you pass null, null for next and previous. Its similar to creating a head as head doesn't have next and previous nodes.
When you create the head you will never change it but you will change the next of the head the time when you create the second element in your list
When you create a tail of the linked list
public void addToTail (int el) {
if (!isEmpty ( )) {
tail = new Node (el, null, tail);
tail.prev.next = tail;
}
else head = tail = new Node(el);
}
here you first create a tail Node by tail = new Node (el, null, tail);
And then get the previous of tail and set the next element of the prev as the tail by doing tail.prev.next = tail;
Every time you add a new Node to the list you are calling addToTail(int e1) which updates the tail and updates the next of the old tail.
Let's suppose the entry point of your code is a call to:
List list = new List();
list.addToTail(1);
list.addToTail(2);
list.removeFromTail();
The first line of code execute the constructor into the list class, which simply set head and tail equal to null.
The second line of code add the method addToTail, and set the variable head and tail equal to new Node(1), which invoke the constructor having one parameter only in the Node class.
The third line of code add the method addToTail, but this time the list is not empty, so the script enters in the if statement and the node that you save with Node(2, null, tail). If you check this Node constructor with three parameters, tail (the node you added before which value was 1) will be set as the prev (previous) node associated with the current one.
The last line of code remove the 2nd node with the code tail = tail.prev; tail.next = null; and return the element you just removed.
firstly node in java its a specific type of variable, which can do the normal work of a variable and have an other information allow her to point on an other node in your case is public Node next, prev;
usually node are used in list ,think like train this is why you have addtootail here or removefromtail .
in this link you find a pretty quite explanation about that.
http://www.vias.org/javacourse/chap14_02.html
The Node class encapsulates the data, in this case an integer value named info.
Also, the List is implemented as a sequence of Nodes.
The reason for the next and prev values in Node is so that the node can be linked to its previous and next nodes in the list sequence. The only way to access a node in the list is to traverse the list, either forwards using the next reference, or backwards using the prev reference.
Note that lists do not support random access like an array would. To get to an element, you need to traverse the list in some direction.
I need to know what's happening behind the scene when the code executes(especially how the line3 works).
Please explain how the Node class works.
public Node next, prev;
Java variables that refer to object types are references, not values. The fields next and prev are not Node objects. No Node object contains other Nodes. Instead, next and prev reference Node objects stored separately.
If you're coming from C or C++, you can think of these references somewhat like pointers.
EDIT:
If you're instead asking about the construction of linked lists in general, your Node class represents a single node, containing a single item, in a doubly linked list. In a doubly linked list, each node refers to the node before and the node after.
See this illustration of a doubly linked list in this related wikipedia article.
The List is like a train. A Node is like a car in the train. The links are references between Nodes, which are similar to the coupling between train cars. The next and prev fields that you ask about are the links between nodes.
Logically, the List contains the entire train. Physically, your List references the head and tail of the train, so that it can travel over the Nodes from front-to-back, or back-to-front.