Implementing a set using a linked list

My task is to add the specified element to the linked list if it is not already present

/**
* Adds the specified element to this set if it is not already present.
* @param o element to be added to this set.
* @return  true if this set did not already contain the specified element.
* @throws NullPointerException - if the specified element is null and this set  does 
  not support null elements
 */

This is my solution

public boolean add(E o)  {

    if (o == null)
        throw new NullPointerException();

    ListNode newNode = new ListNode(o, null);
    if (contains(o))
    {
     return false;
    }
    else {

    head.next=newNode;
    head=newNode;
    return true;
   }     
}

Can please anybody check it and correct if there are mistakes, or give hints to correct it?

Let's say that you are starting with a list containing five elements:

$$
\newcommand{ptr}[1]{\overset{\mathtt{#1}}{\longrightarrow}}
\mathtt{head} \ptr{} \fbox{first} \ptr{next}
\fbox{second} \ptr{next}
\fbox{rest} \ptr{next}
\fbox{of} \ptr{next}
\fbox{list} \ptr{next} \mathtt{null}
$$

Then, you add a non-duplicate item called "another". After executing head.next=newNode:

$$
% "align" allows us to do vertical alignment
% "\\" separates lines
% "&" is an alignment marker. The markers in different rows
% are positioned above each other, and horizontally centered.
\newcommand{ptr}[1]{\overset{\mathtt{#1}}{\longrightarrow}}
\begin{align*}
&\mathtt{newNode} \\
&\quad\downarrow \\
\mathtt{head} \ptr{} \fbox{first} \ptr{next} &\fbox{another} \ptr{next} \mathtt{null} \\
&\fbox{second} \ptr{next}
\fbox{rest} \ptr{next}
\fbox{of} \ptr{next}
\fbox{list} \ptr{next}
\mathtt{null}
\end{align*}
$$

You've just orphaned and discarded everything starting from the second element. Then it gets worse when you set head=newNode.

StackExchange Code Review Q#48679, answer score: 11