Get Node Value [HackerRank solution]

You’re given the pointer to the head node of a linked list and a specific position. Counting backwards from the tail node of the linked list, get the value of the node at the given position. A position of 0 corresponds to the tail, 1 corresponds to the node before the tail and so on.

Input Format
You have to complete the int GetNode(Node* head, int positionFromTail) method which takes two arguments – the head of the linked list and the position of the node from the tail. positionFromTail will be at least 0 and less than the number of nodes in the list. You should NOT read any input from stdin/console.

Constraints
Position will be a valid element in linked list.

Output Format
Find the node at the given position counting backwards from the tail. Then return the data contained in this node. Do NOT print anything to stdout/console.

Sample Input

1 -> 3 -> 5 -> 6 -> NULL, positionFromTail = 0
1 -> 3 -> 5 -> 6 -> NULL, positionFromTail = 2

Sample Output

6
3

 CODE:

int GetNode(Node *head,int positionFromTail)
{
 // This is a "method-only" submission. 
 // You only need to complete this method. 
 int a[1000];
 int value;
 int i=0;
 while(head)
 {
 a[i] = head->data;
 i++;
 head = head->next;
 }
 value = i - positionFromTail-1;
 return a[value];
}

// PASSED ALL TEST CASES.