how does this work? there is no link between the 2ndlast item (formerly the last) and the new one just added. Also current.next, when current is null woud return an NPE, wouldnt it?

Please suggest , unable to find out mistake in this :

Node* Insert(Node *head,int data)
{
Node *temp;
temp->data = data;
temp->next = NULL;

if(head == NULL)
  {
  head = temp;
}
else
  {
  Node *p;
  p = head;
  while(p->next != NULL)
      p = p->next;
  p->next = temp;
}
return(head);
}

Showing no response in output :(

In your code:

if(head == null) {
    head = tmp;
    return head;
} 

Why not simply return tmp instead of re-assigning head to tmp and then returning head?

Just curious. Thanks!

thanku it helped a lot ...

Below approach is with recursion

Node Insert(Node head,int data) {
    return insertDS(head, data, head);
}
Node insertDS(Node head, int data, Node parent){
	if(head == null){
		head = new Node();
		head.data = data;
		if(parent != null) parent.next = head;
	}else
		insertDS(head.next, data, head);
	return head;
}

this doesn't work

it's not working

var newNode = new SinglyLinkedListNode(data); if(head == null){ head = newNode; return head; }else{ var temp = head; while(temp.next!=null){ temp = temp.next; } temp.next = newNode; return head }

Hope this will helps you !! :)

• var newNode = new SinglyLinkedListNode(data);
• if(head == null){
• head = newNode;
• return head;
• }else{
• var temp = head;
• while(temp.next!=null){
• temp = temp.next;
• }
• temp.next = newNode;
• return head
• }*

Thank you! In the 'else' case, I was returning the 'Insert' result and I see why that is incorrect now.

In the 3rd line you return a new Node object. I thought we were supposed to return only the head object. Can you explain this part? My understanding with linked list is that the head never has any data but is used to reference the first element in the list, yet you return a single Node with data and no head.

Nice use of the floating recursive call.

nice code

I did it non recursively, calling a function multiple times unless necessary would be inefficient.

def Insert(head, data):
    if head == None:
        return Node(data,None)
    temp = head
    while temp != None:
        prevFinalNode=temp
        temp = temp.next
    prevFinalNode.next = Node(data,None)
    return head

Also you could have used elif instead of else>if/else as such

def Insert(head, data):
    if head == None:
        return Node(data, None)
    elif head.next == None:
        head.next = Node(data,None)
    else:
        Insert(head.next,data)
    return head

The code looks more readable this way I think, sorry for the nitpicking.

Nice! Good to see how to solve this problem in different ways.

But Would you please clear one doubt! How is head able to use next attribute? Node class object only can have it, right ?

why u returned head ? in last line u can return head.next ?

Hey,

If we go in the else part, we should return current instead of head right?

How does returning head give us the correct solution? Can you please explain?

my program is the same as yours except the first part of the code where mine says

Node *temp;    //did not use Node*temp=new Node
temp->data=data;
temp->next=NULL;

This is causing error in my code.. Can you please explain?