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

To understand this problem, we should do a simple analogy with a measurement example. Let's say, you have to find the place of your arm where exactly 1 meter away from your middle finger, how would you measure? You would just grab a ruler with a 1-meter length and put the top-end of that ruler to the tip of your middle-finger and the bottom-end of the meter will be exactly 1 meter away from the top of your middle-finger.

What we do in this example will be the same, we just need a frame with n-element wide and what we have to do is put the frame to the end of the list, thus the start node of the frame will be exactly n-th element to the end of the list.

This is our list assuming we have M elements in the list, and our frame with N element wide;

HEAD -> EL(1) -> EL(2) -> ... -> EL(M-1) -> EL(M)

<-- Frame -->

However, we only need the boundaries of the frame, thus the end boundary of the frame will exactly (N-1) elements away from the start boundary of the frame. So have to only store these boundary elements. Let's call them A and B;

HEAD -> EL(1) -> EL(2) -> ... -> EL(M-1) -> EL(M)

A <- N-Element Wide-> B

The first thing we have to do is finding B, which is the end of the frame.

ListNode b = head;
int count = 1;

while(count < n && b != null) {
    b = b.next;
    count++;
}

Now b is the n-th element of the array, and a is located on the HEAD. So our frame is set, what we will do is increment both boundary nodes step by step until b reachs to the end of the list where a will be n-th-to-the-last element;

ListNode a = head;

while(b.next != null) {
    a = a.next;
    b = b.next;
}

return a;

To gather up everything, and with the HEAD checks, N < M (where M is the size of the list) checks and other stuff, here is the complete solution method;

public ListNode findNthToLast(int n) {
    if(head == null) {
        return null;
    } else {
        ListNode b = head;
        int count = 1;

        while(count < n && b != null) {
            b = b.next;
            count++;
        }

        if(count == n && b!=null) {
            ListNode a = head;

            while(b.next != null) {
                a = a.next;
                b = b.next;
            }

            return a;
        } else {
            System.out.print("N(" + n + ") must be equal or smaller then the size of the list");
            return null;
        }
    }
}