Exam 2 Answers
Section 02

1. D
2. E
3. B
4. B
5. False

6.    2
      3
      10
      26
   

7. 

8.    public void setNth (int loc, T value) {
         if (loc > numElements -1)
            throw new QueueUnderflowException();
         int i = 0;
         T hold = null;
         LLNode<T> curr = front;
         for (i = 0 ; i < loc ; i++)
            curr = curr.getLink();
         curr.setInfo(value);
      }
   

9. The setNth method will have to start at the beginning of the linked list and access loc + 1 elements to find and return the element at position loc. In the worst case loc is N-1, which gives a complexity of O(N).

10. 1.  

       
    2. This compareTo method is in the Printer class:

          public int compareTo(Printer printer) {
             int eq = manufact.compareTo(printer.manufact);
             if (eq != 0)
                return eq;
             return model.compareTo(printer.model);
          }
       

    3.    SortedABList<Printer> printersSorted = new SortedABList<>();
       

    4. Like the compareTo method, this method is in the Printer class.

    5.    SortedABList<Album> speedSorted = new SortedABList<>(Printer.speedComparator());
       

    6.    public int countAfter(ABList<Printer> prList, Printer printer) {
            int count = 0;
            Printer prFromList;
            Iterator<Printer> prIter = prList.iterator();
            while (prIter.hasNext()) {
               prFromList = prIter.next();
               if (prFromList.compareTo(printer) > 0)
                  count++;
            }
            return count;
          }
       

11. 1.    public void enqueue(T newItem) {
             elements.add(elements.size(), newItem);
          }
       

    2.    public T dequeue() {
             if (elements.size() == 0)
                throw new QueueUndeflowException();
             T item = elements.get(0);
             elements.remove(0);
             return item;
          }
       

12. 1. No, it is not balanced. It's out of balance at 470.
    2. No, it's not a BST because 365 is in the right subtree of 375 and 485 is in the left subtree of 470.
    3. The height is equal to the maximum level of a node. The maximum level is 4 (nodes 199, 432, and 485) so the height is 4.
    4. pre: 375 255 170 109 233 199 282 303 470 417 365 461 432 485 566
       in: 109 170 199 233 255 282 303 375 365 417 432 461 485 470 566
       post: 109 199 233 170 303 282 255 365 432 485 461 417 566 470 375

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