# Get nth node from end in a linked list

Description:

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.

Code:

int GetNode(Node head,int n) {
// This is a "method-only" submission.
// You only need to complete this method.
int count = 0;
while (current != null) {
count++;
current = current.next;
}
for (int i = 0; i < count - n - 1; i++) { // extra -1 to avoid going out of linked list.
current = current.next;
}
return current.data;
}

• What if you are passed an empty list (head is null)? How do you want to handle the case of n >= the length of the list? Your code returns the first list element, @janos code will simply fail. In an assignment I would at least want to see that you realized there are potential problems. Dec 23, 2016 at 11:57

Instead of a count variable and reusing current, I think it's neater to use two pointers:

Node runner = head;
for (int i = 0; i < n; i++) {
runner = runner.next;
}
while (runner.next != null) {
runner = runner.next;
current = current.next;
}
return current.data;


Other than that, your implementation is fine, except for a few tiny style points that are barely worth mentioning, but here we go anyway:

• The commented out instructions about method-only submissions are unnecessary
• Add a space after commas in parameter list
• I have to agree, a very neat solution indeed. I wonder how to get thinking like that :) Dec 23, 2016 at 6:33
• Me too :-) I saw it once and can't forget about it. Dec 23, 2016 at 6:34
• Implementation in C, Java and Python: geeksforgeeks.org/nth-node-from-the-end-of-a-linked-list Dec 23, 2016 at 6:43