How to find nth element from the end of a singly linked list?

Question

The following function is trying to find the nth to last element of a singly linked list.

For example:

If the elements are

Answer 1

There are lots of answers here already, but they all walk the list twice (either sequentially or in parallel) or use a lot of extra storage.

You can do this while walking the list just once (plus a little bit) using constant extra space:

Node *getNthFromEnd(Node *list, int n) {

    if (list == null || n<1) {
        return null; //no such element
    }

    Node *mark1 = list, *mark2 = list, *markend = list;
    int pos1 = 0, pos2 = 0, posend = 0;

    while (markend!=null) {
        if ((posend-pos2)>=(n-1)) {
            mark1=mark2;
            pos1=pos2;
            mark2=markend;
            pos2=posend;
        }
        markend=markend->next;
        ++posend;
    }
    if (posend n) {
      mark1 = mark1->next;
      ++pos1;
    }
    return mark1;
}

This version uses 2 extra pointers does less than N+n traversals, where N is the length of the list and n is the argument.

If you use M extra pointers, you can get that down to N+ceil(n/(M-1)) (and you should store them in a circular buffer)