EPI: Zip of a linked list

http://programming-puzzle.blogspot.com/2014/02/zip-of-linked-list.html
Given a linked list <1, 2, 3, 4, 5, 6>, zip of this linked list is defined as 1, 6 , 2, 5 , 3, 4. And the task is to achieve desired linked list using O(1) space. 
This can be performed by a simple algorithm:

• Split the list from the middle into two lists. We are splitting the list into two and not creating a new linked list hence maintaining O(1) space
• Now we have two lists : 1, 2, 3 and 4, 5, 6. Reverse the second list
• This gives us two lists 1, 2, 3 and 6, 5, 4
• Now merge the lists picking one node from each list as a time

Solution from EPI
  public static <T> Node<T> zippingLinkedList(Node<T> L) {
    Node<T> slow = L, fast = L, preSlow = null;

    // Find the middle point of L.
    while (fast != null) {
      fast = fast.next;
      if (fast != null) {
        preSlow = slow;
        fast = fast.next;
        slow = slow.next;
      }
    }

    if (preSlow == null) {
      return L; // only contains one node in the list.
    }
    preSlow.next = null; // splits the list into two lists.
    Node<T> reverse = ReverseLinkedListIterativeTemplate
        .reverseLinkedList(slow);
    Node<T> curr = L;

    // Zipping the list.
    while (curr != null && reverse != null) {
      Node<T> temp = curr.next;
      curr.next = reverse;
      curr = temp;
      // Connect curr->next to reverse, and advance curr.
      // connectANextToBAdvanceA(ref_curr, reverse);
      if (curr != null) {
        // Connect reverse->next to curr, and advance reverse.
        Node<T> temp2 = reverse.next;
        reverse.next = curr;
        reverse = temp2;
        // connectANextToBAdvanceA(ref_reverse, curr);
      }
    }

    return L;
  }