CareerCup

You can maintain three variable to store the top 3 elements while sweeping the list.

1. Start one pointer "fp" and iterate through linkedlist until it reaches 3rd node.
2. When "fp" reaches 3rd element, start another pointer "sp" from head node.
3. From now on, advance both pointers"fp" and "sp", node by node until you reach "fp"pointing to NULL.
4. At this time "sp" is pointing to 3rd element from end of the linked list
5. In other words, iterating a pointer to maintain some gap (here 3 in this example)

check if the node->next->next->next is null..if so, then you are at the n-3 rd element.

I don't think you need 3 temps, 2 are sufficient, one t1 = head->next->next->next, and t2 = head.
Make ++t1, and ++t2, until t1 == NULL.

public Link find(Link first)
{
Link current =first;
Link fast=first;
while (fast !=null)
{
fast=fast.next.next.next;
current=current.next;
}
return current;
}

Hi

Node* findNthToLast(Node* head, int n)
{
	Node* temp = head;
	Node* nthToLast = NULL;
	int count = 0;
	while(temp != NULL)
		{
			count++;
			temp = temp->next
			if(count == n)
			{
				nthToLast = head;
			}
			else if (count > n)
			{
				nthToLast = nthToLast ->next;
			}

		}
		if(nthToLast == NULL)
		{
			cout << "Error: List size is smaller than N"<<endl;
		}
		return nthToLast;
}

public void getNthMinusyElement(int n){
System.out.println();
ListNode curr = head;
ListNode nthToLast = null;
int count = 0;
while(curr != null)
{
count++;
curr = curr.getNext();
if(count == n)
{
nthToLast = head;
}
else if (count > n)
{
nthToLast = nthToLast.getNext();
System.out.println("data==="+nthToLast.getData());

}

}
System.out.println(nthToLast.getData());

}