CODEWITHSUNDEEP
javaalgorithmooplinked-list
hussein albirouti
hussein albirouti

Reputation: 127

which role each Node does it play in linked list?

Why each time we create new node p.next so we need to assign null to this.next? isn't always null any way? and which role does it play in LinkedList ? if we try to print out this.next it will be null before we assign it to null.

System.out.println(this.next);

Result is null

Also if p.next point to the new node why we need to set p = p.next to point to the same node? if the purpose to set tail to p at last, can't we just set tail = p.next which is the last one after for loop is finished.

public class EnkeltLenketListe<T>{
private T value;
private Node<T> next;

private Node(T value, Node<T> next)
{
System.out.println(this.next);
    this.next = next;
    this.value = value;
}
}

private Node<T> head, tail;

public EnkeltLenketListe(T[] a)
{
this();

int i = 0; for (; i < a.length && a[i] == null; i++);

if (i < a.length)
{
    head = new Node<>(a[i], null);
    Node<T> p = head;

    for (i++; i < a.length; i++)
    {
        if (a[i] != null)
        {
            p.next = new Node<>(a[i], null);
            p = p.next;
        }
    }
    tail = p;
}
}

Upvotes: 3

Views: 92

Answers (2)

The Scientific Method
The Scientific Method

Reputation: 2436

Why each time we create new node p.next so we need to assign null to this.next?

while adding and removing a nodes, we have to make sure that nodes don't point to unintended nodes. they might have pointing to some nodes.

why we need to set p = p.next to point to the same node

to locate and maintain your position in the list while you are traversing the list, you start from head, and continue to remaining nodes. if p=p.next, how are going to traverse the list?

can't we just set tail = p.next which is the last one after for loop is finished.

No, we can not because in this case p.next is equivalent to p.p.next because p was set to p.next inside the loop. test it by adding the following before tail=p, you should get null

System.out.println(p.next);


Edited:

your list is singly linked list which means every node except the tail should have a pointer to next node, you started with

head = new Node<>(a[i], null);
Node<T> p = head;

in this case p and head are pointing Node 0 see image below. if the next NOde in the array is not null, let see what happens in the for loop

    p.next = new Node<>(a[i], null);
    p = p.next;

In this case p.next is pointing to Node 1 (see image below), where as p which were pointing to Node 0 is now set to point to Node 1. so both are pointing to 'Node 1`. the last one:

tail = p;

You said that why don't we just tail=p.next? No, we can not because in this case p.next is equivalent to p.p.next because p was set to p.next inside the loop.

singly linked list

read about singly linked list here

Upvotes: 3

OmG
OmG

Reputation: 18838

Try to use contradiction. If you didn't set p = p.next, in the next iteration of the loop, again you will set the new node to the next position of the previous node. Hence, all the time p is the head and you didn't move the p anymore!

Therefore, you need to move p in each iteration after setting p.next. Indeed, it is a pointer to the last element of the list.

Upvotes: 3

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