CareerCup

-use two pointers
-move one of the pointers n positions
-now move both the pointers simultaneously till the second pointer get's null
-ur first pointer will point to nth last element

I only know Java, so here it is.

public class LinkedLists {
    static class Node {
        String key;
        Node   next;
    }

    public static void main(String[] args) {
        System.out.println(findNthLast(createList("a"), 1).key);
        System.out.println(findNthLast(createList("a b"), 2).key);
        System.out.println(findNthLast(createList("a b c"), 2).key);
        System.out.println(findNthLast(createList("a b c d"), 2).key);
        System.out.println(findNthLast(createList("a b c d e"), 2).key);
        System.out.println(findNthLast(createList("a b c d e f"), 2).key);
    }

    public static Node createList(String keys) {
        String[] parts = keys.split("\\s+");
        Node list = new Node();
        Node node = list;
        node.key = parts[0];
        for (int i = 1; i < parts.length; i++) {
            node.next = new Node();
            node.next.key = parts[i];
            node = node.next;
        }
        return list;
    }

    public static Node findNthLast(Node list, int n) {
        int i = 1;
        Node node = list;
        Node nthLast = null;
        while (node != null) {
            if (i == n) {
                nthLast = list;
            }
            if (i > n) {
                nthLast = nthLast.next;
            }
            node = node.next;
            i++;
        }

        return nthLast;
    }
}

I tried a simple algorithm and implemented in c++

kodeyard.blogspot.com/2011/09/nth-last-element-in-linked-list-c.html

FindNthLast(node *l, int n, node **ans)
{
if(l -> next == NULL) return 1;
p = FindNthLast(l->next, n, ans);
if( p == n )
*ans = l;
return p + 1;
}