Why use "node.getNext( ) == null" as a convention for defunct node? - java

This is a code fragment from the book "Data Structures and Algorithms in Java (6th Ed.)" This is method is part of the LinkedPositionalList implementation.
I don't understand exactly what does the line:
"if (node.getNext( ) == null) // convention for defunct node"
is for. I'd appreciate some explanation about its functionality.
// private utilities
/** Validates the position and returns it as a node. */
private Node<E> validate(Position<E> p) throws IllegalArgumentException {
if (!(p instanceof Node)) throw new IllegalArgumentException("Invalid p");
Node<E> node = (Node<E>) p; // safe cast
if (node.getNext( ) == null) // convention for defunct node
throw new IllegalArgumentException("p is no longer in the list");
return node;
}

The implementation in the book you're asking about uses an explicit trailer node that represents a "one past the end" position in the structure. Therefore, all nodes in the list must have a next node. Therefore, if a node's getNext() method returns null that node isn't in the list.

In this case defunct node means a non existent node.
if (node.getNext( ) == null) // convention for defunct node
In this case, the node p has a method called getNext() which can return null, if it does, then it means its the last of the nodes. In other words there is no next node.
This kind of structures, as the name implies, link between each other. getNext() should return a position in memory where the next node is, if such position is not existent then, there is no node there.
Usually node will likely have a method called setNext(Position<E> next) this is the counterpart that saves the position of the next item for this node.

If you read in the book the "Doubly Linked list Implementation" part of chapter 7 explains the defunct position notion as a result of getting rid of any removed position (node) in the list. Because the remove() method implements that the position to be deleted is set to null value.
So the validate() method actually checks if the input position is not removed from the list.. how? once again, by checking if calling getNext() on it would get null value. If so, then it must have been set to null as a deleted position using the remove() method. Therefore it throws an error as the input position p is no longer in the list (since it's a deleted position).
To better understand, just refer to the delete() method there in the book. Here it is from the book:
// Removes the element stored at Position p and returns it (invalidating p).
public E remove(Position<E> p) throws IllegalArgumentException {
Node<E> node = validate(p);
Node<E> predecessor = node.getPrev();
Node<E> successor = node.getNext();
predecessor.setNext(successor);
successor.setPrev(predecessor);
size−−;
E answer = node.getElement();
node.setElement(null); // help with garbage collection
node.setNext(null); // and convention for defunct node
node.setPrev(null);
return answer;
}

Related

Why we can use Node head = null without instantiate the 'head' in Java?

I am reading someone's code. It is about getting input numbers and convert those number into a Linked list. The part of the code is like this:
class Node {
int value;
Node next;
Node() {
next = null;
}
}
Firstly We need to create a head node to indicate head and we let the head be null like this Node head = null.
My limited experiences of java tell me that head is supposed to be a Node type object here. So why we can use Node head = null without instantiate the head?
I think at least I should create Node head = new Node(); then we can use Node head = null;
Anyone can explain it to me?
Node head = null;
This line states that there are no items in the linked list. This is valid in Java and indicates that although head can contain a Node object (or an object of a derived class), it is currently empty.
To add an item to the list, there is likely some code such as:
public void addItemToLinkedList(int value) {
if (head == null) {
head = new Node();
head.value = value;
} else {
...
}
}
So if there is no first Node (when head equals null) then head is created. Otherwise if head already exists, the else block would execute which would look for the end of the list and add another Node.
head is supposed to be a Node type object here
This is optional. Java allows head to be a Node object, or null, to indicate that head is not referencing any nodes at all.
The value of null is special in Java. It is allowed to go wherever an object can go. It indicates that the variable to which you assign null is empty. This is perfectly fine, assuming that the rest of your program deals with null checking.

What does "this" refer to in Linked List example?

So I'm skimming through Cracking the Coding Interview to brush up on some interview stuff and I ran across this linked list implementation, and maybe it's been a while but it's completely going over my head. I understand most of it, except for one specific line, and it's throwing me off. I'll post the code below (for reference, the book doesn't mention language but it appears to be Java.)
class Node {
Node next = null;
int data;
public Node(int d) {
data = d;
}
void appendToTail(int d) {
Node end = new Node(d);
Node n = this;
while(n.next != null) {
n = n.next;
}
n.next = end;
}
}
I'm a little confused on the line: Node n = this - I'm not sure what this is referring to, unless it's talking about next - why not just set it to null in that case?
this refers to a specific instance of an object of a class. Since objects are constructed there can be multiple instances of a class, but using the this keyword allows you to obtain a reference to itself, meaning a reference to the the specific instance of the object whose method is being called.
The linked list is a collection of nodes that are, well, linked together. When you call appendToTail() the node will look at all of the Node objects linked to itself and follow the chain. For it to get a reference to itself to follow its own chain the this keyword is used.
You also ask why null isn't used in this case to initialize n. This would cause a NullPointerException when n.next is first called in the loop constraint, so instead its own reference is used as the starting point for the iteration of the linked-list.
This (pun intended) can be a confusing topic at first, but lets use the example you provided.
Node n = this;
while(n.next != null) {
n = n.next;
}
Let's pretend that there are 4 objects currently linked in our list and for simplicity's sake the Node object that appendToTail() is being called on is the head of the list. Here's the reference value of Node n that's held on each loop iteration from the above snippet.
We're pointing to ourself - this
Pointing to the second item in the linked list. - this.next
Pointing to the following item - this.next.next
Pointing to the last item in the list - this.next.next.next
The loop ended so currently the reference of n = this.next.next.next. We then set n's next value (where n is currently pointing to the end of the linked chain) to the new object we created at the beginning of our method, which makes it the new end of the list. (n.next = end is now equivalent to this.next.next.next.next = end).
Semi-Unnecessary Edit: This is explained in terms of Java. It appears that someone added the C++ tag after I wrote this answer
This is Java.
"this" refers to the specific instance of the class in which the call is being made. In this case, "this" is in reference to the specific class Node you are dealing with. While the variable "end" creates a new and separate version of the Node class which is constructed using the passed int "d".
Since this is a Linked List all Nodes are connected and you have a start Node (root). So when using it it would look like this:
Node root = new Node(6); //need an instance first
root.appendToTail(5);
root.appendToTail(3);
//6->5->3
Since this the nodes are connected I need one start Node and need to check if this has a next node when yes I need to search deeper. When a node did not have a next Node it is the current last one and can add my new Node. So this in Java refers to the current instance of a class. In my example the root Node(because I call root.appendToTail). So the method will search from the root Node (value 6) the next Node without a next Node (the one with value 3) and append it there. If I can get a child reference and would call child3.appendToTail the method would search from child3 instead of starting from my root.
When setting n to null and rewriting the while to go from this.next you would have a problem when the current node you use appendToTail did not have a next Node and an NullPointerException would be thrown.
Node n = this; means n object references to the object which is calling this method.
So method is looping to next object till next object is null and assigning end node to the end.
Lets see
1 -- 2 -- 3 -- 4
*
|
*
obj
you have an obj object that is pointing to node 1. When u call obj.appendToTail(5)
Node end = new Node(d); //new node is created to add to the end.
Node n = this; //local n object is referenced to node 1(or obj)
while(n.next != null) {
n = n.next;
}
//n here is node 4 since there is no next node to 4
n.next = end; //node 5 is tail now
End result:
1 -- 2 -- 3 -- 4 -- 5
As you can see in this code
class Node {
//
void appendToTail( int d ) {
Node *end = new Node( d );
Node n = this;
// ...
}
}
Your class Node has a reference to a Node in it's definition.
The line: Node *end = new Node( d ); means inside a Node there is a reference to another node.
The line Node n = this; means inside a Node the reference to that node itself, is represented by this. Ergo, n is also a reference to said node itself.
Any Node instance can call appendToTail().
Notice howebet, here in fact Node does not append itself to tail of the list, what happens here is that new node is created and added to tail, not the one on which method is invoked.
For this to happen, first we need to find the tail of the list given the current Node.
// n is pointing to current Node
while(n.next != null) {
n = n.next;
}
Once we find node which has next == null, this is tail of the list so we can now append new Node to the tail:
// n points to current tail before next line is invoked
n.next = end;
As to why there is line:
Node n = this;
Well since there is no LinkedList class which maintains reference to the head, you have to be able to iterate from any given Node. That is what happens here, you start iteration from Node on which appendToTail is called, but that Node can be anything at this point, from head to tail.
As a side note, if you implement Linked List by hand, make sure to actually have class LinkedList which will offer methods such as add, get, size, appendToTail and so on, rather then putting these into Node class.

Implementation of removeFirst() method in SLinkedList in java

I got the following code from one book for implementing a singly linked list. And I don't understand some lines of code in the removeFirst() method, which removes the first node from the LinkedList.
class ListNode{
private String element;
private ListNode next;
public ListNode(){
element = null;
next = null;
}
public ListNode(String s, ListNode n){
element = s;
next = n;
}
//Access method
public String getElement(){
return element;
}
public ListNode getNext(){
return next;
}
//Modify method
public void setNext(ListNode n){
next = n;
}
}
public String removeFirst(){
if(head == null)
return null;
else{
ListNode temp = head;
head = head.getNext();
temp.setNext(null); //Which I don't understand, is it necessary?
size --;
return temp.getElement();
}
}
It seems that the statement temp.setNext(null); can be omitted. So why it is here, does it has anything to do with the garbage colletion in java. Since I am new to Java, any suggestions or ideas?
It depends on the entire implementation of the linked list, which you have not included in your question. However if it is possible for objects to hold a reference to a node even after if has been removed from the list, then the line is necessary.
Suppose we have a long chain of nodes A -> B -> C -> .... Suppose all of these nodes have been removed from the list, but that we still hold onto a reference to A. If all the nodes still held a reference to the next, this would prevent all of the nodes from being garbage collected. Simply setting the next node to be null ensures that only A cannot be garbage collected.
It is likely that implementations of a linked list do mean that references to nodes can be retained. For example, many implementations of Iterator hold a reference to the current node.
Consider this code:
Iterator<String> iterator = list.iterator();
while (i.hasNext()) {
if ("foo".equals(i.next())) {
i.remove();
break;
}
}
// lots more code
This code searches a list for the first occurrence of the String "foo". If it is found, it removes the "foo" from the list and breaks from the loop. The trouble with this is that the Iterator i is still in scope for the remaining code and still holds a reference to a node. This node may be in the middle of the list if the break occurred. Without setting next to be null, this would prevent all subsequent nodes from being garbage collected while i is still in scope, even if the list is cleared.
Note that you should generally make an iterator local to a loop anyway, like this
for (Iterator<String> i = list.iterator();;i.hasNext())
Let us suppose, single linked list of nodes is : A --> B --> C --> D with head node as A. Now, lets go though your removeFirst() method. When it get called, if(head == null) condition doesnt satisfy because our head node "A" is not NULL. Then it execute else statement,
temp = head //take head reference into temporary variable
head = head.getNext(); //As we want to delete first Node which is head so we are setting head to next node (i.e. head --> next) which is Node "B"
Now, to delete A (first Node) we need to break connection between A (which is previous head)--> B (current head) and that will done by
temp.setNext(null); //so out linked list became A B-->C-->D
and then as one node is deleted so in next statement it decreases size of link by size--.
I think we also set the temp reference to NULL as temp=NULL to make deleted node eligible for garbage collection.

Remove recursively from a binary search tree

This is homework; please don't just give me code
I have two methods: remove(T data) and removeRec(Node<T> node, T data).
In its current state, it seems my code only removes the root node of the BST.
#Override
public T remove(T data) {
if (data == null) {
throw new IllegalArgumentException("Data is null");
}
if (root == null) {
throw new java.util.NoSuchElementException("BST is empty");
} else {
size--;
BSTNode<T> dummy = new BSTNode<T>(null);
return removeRec(root, data, dummy).getData(); //This is probably wrong too
}
}
/**
* Helper method to recursively search for, and remove the BSTNode with
* the given data in it
* #param node is the node we're currently at
* #param data is the data we're looking for
* #param temp I have no idea why
* #return node that was removed
*/
private BSTNode<T> removeRec(BSTNode<T> node, T data, BSTNode<T> temp) {
if (compare(data, node.getData()) < 0) {
temp.setLeft(removeRec(node.getLeft(), data, temp));
} else if (compare(data, node.getData()) > 0) {
temp.setRight(removeRec(node.getRight(), data, temp));
} else if (node.getLeft() != null && node.getRight() != null) {
temp.setData(findMin(node.getRight()).getData());
temp.setRight(removeRec(node.getRight(), data, temp));
} else {
if (node.getLeft() != null) {
temp = node.getLeft();
} else {
temp = node.getRight();
}
}
return temp;
}
private int compare(T a, T b) {
return a.compareTo(b);
}
My instructor has told me (as a hint) that I should see what passing in a third argument into the method, in this case, BSTNode<T> temp. I don't understand how that helps though, or how to utilize it. I don't see how using a third argument helps; and I can't find anything online as to why you'd do this either.
There are three main possibilities when you try to remove data from your Binary Search Tree:
data is less than the current node value: Call remove on the left subtree or throw a NoSuchElementException if it is null.
data is greater than the current node value: Call remove on the right subtree or throw a NoSuchElementException if it is null.
data is equal to the current node value.
1 and 2 are pretty straightforward, but 3 has four more cases to consider:
3.1. current node is a leaf: Both left and right subtrees are null. Just replace the reference to the current node in its parent by null.
3.2. current node has only the left child: You need to make the parent of the current node point to the left subtree, thus removing the current point. To do this, you can implement a function that will check if the current point was on the left or right subtree of the parent and replace it accordingly. Calling it would look like this:
replaceNodeInParent(node, node.getLeft(), parent);
3.3. current node has only the right child: Similar to 3.4, but using getRight() instead of getLeft().
3.4. current node has both the left and right children: You should maintain the property of the BST that all nodes on the left are less than the current node and all nodes on the right are greater than the current node. To do so, you should find the smallest value on the right, copy it to the current node, and delete it from the right subtree. Something like this:
BSTNode<T> successor = findMin(node.getRight());
node.setData(successor.getData());
removeRec(node.getRight(), successor.getData(), node);
It looks like your BSTNode doesn't hold a reference to the parent node. If so, I believe that's what the third argument for removeRec should be. You will need a reference to the parent every time you replace the current node, so you can set the parent left or right subtree as needed.
For further reading, you can check this article on Binary Search Trees from Wikipedia.

Setting myself to null - Java

I came across the following problem:
Delete a node in the middle of a singly linked list, given only access to that node. (head is not given)
Now there are a lot of solutions and they all do not work when the element to be deleted is the last node.
Why wouldn't this work?
public static void removeNode (Node n){
if(n.next == null){ //n is the last node
n= null;
return;
}
//handling general case here
}
Java passes parameters by value, so setting n to null has no effect outside of the method. This means the method essentially does nothing when passed the last node of a list.
You need to set null the reference in the previous node, not the variable that references to your last node, something like this:
if(n.next == null) {
prev.next = null;
return;
}
n is local to the method, so changing its value won't affect the list itself. You need to modify the next of the previous node, which you do not have access to.

Categories