Implementing find middle element of singly linked list without using 2 pointers

Question

There is a solution floating around about using 2 pointers. But I decide to keep the implementation much simpler by using basic constructs that works use to build basic methods of a singly list list data structure.

The code works. My question is, am I off the hook from making it better as long as the code works? Is the code sufficient?

 public Object findMiddleElement()
 {
 Node node = head;
 if(size % 2 != 0)
 {
 for(int i = 0; i < size/2 ;i++ )
 {
 node = node.getNext();
 }
 }
 else
 {
 System.out.println("there is no middle element");
 node.setElement(null);;
 }
 return node.getElement();
 }

Answer 1

It could be a little easier to read like this:

public Object findMiddleElement() {
 Node node = head;
 if(hasMiddleElement()) {
 for(int i = 0; i < size / 2; i++) {
 node = node.getNext();
 }
 } else {
 System.out.println("there is no middle element");
 node.setElement(null);
 }
 return node.getElement();
}
private boolean hasMiddleElement() {
 return size % 2 == 1;
}

This works if size is set in the class.