Nth node from end of linked list.cpp

/*
Nth node from end of linked list
================================

Given a linked list consisting of L nodes and given a number N. The task is to find the Nth node from the end of the linked list.

Example 1:
Input:
N = 2
LinkedList: 1->2->3->4->5->6->7->8->9
Output: 8
Explanation: In the first example, there
are 9 nodes in linked list and we need
to find 2nd node from end. 2nd node
from end os 8.  

Example 2:
Input:
N = 5
LinkedList: 10->5->100->5
Output: -1
Explanation: In the second example, there
are 4 nodes in the linked list and we
need to find 5th from the end. Since 'n'
is more than the number of nodes in the
linked list, the output is -1.
Your Task:
The task is to complete the function getNthFromLast() which takes two arguments: reference to head and N and you need to return Nth from the end or -1 in case node doesn't exist..

Note:
Try to solve in single traversal.

Expected Time Complexity: O(N).
Expected Auxiliary Space: O(1).

Constraints:
1 <= L <= 103
1 <= N <= 103
*/

int getNthFromLast(Node *head, int n)
{
  Node *main_ptr = head;
  Node *ref_ptr = head;

  int count = 0;
  if (head != NULL)
  {
    while (count < n)
    {
      if (ref_ptr == NULL)
        return -1;
      ref_ptr = ref_ptr->next;
      count++;
    }

    if (ref_ptr == NULL)
    {
      head = head->next;
      if (head != NULL)
        return (main_ptr->data);
    }
    else
    {
      while (ref_ptr != NULL)
      {
        main_ptr = main_ptr->next;
        ref_ptr = ref_ptr->next;
      }
      return (main_ptr->data);
    }
  }
  return -1;
}