1. Write the output generated by the following code. (8pts) 2. Write the output generated by the following code (8pts)

Transcription

1 LinkedStack<E> implements OurStack<E> LinkedQueue<E> implements OurQueue<E> ArrayList<E> LinkedList<E> implements List<E> public interface <E> { // Check if the stack is empty to help avoid popping an empty stack. boolean (); // Put element on "top" of this Stack object. void (E element); // Return reference to the element at the top of this stack. E () throws EmptyStackException; // Remove element at top of stack and return a reference to it. E () throws EmptyStackException; public interface <E> { // Return true if this queue has 0 elements boolean (); // Store a reference to any object at the end void (E newel); // Return a reference to the object at the front of this queue E (); // Remove a reference to the element at the front of this Queue and return that reference E (); public interface <E> { // Appends the specified element to the end of this list boolean (E element) // Returns true if this list contains the specified element boolean (E element) // Returns the element at the specified index in this list E (int index) // Removes the element at the specified position in this list boolean (E element) // Returns the number of elements in this list. int () 1. Write the output generated by the following code. (8pts) OurStack<Character> s = new LinkedStack<Character>(); s.push('3'); s.push('2'); System.out.println(s.peek()); System.out.println(s.pop()); System.out.println(s.isEmpty()); 2. Write the output generated by the following code (8pts) OurQueue<String> q = new LinkedQueue<String>(); q.enqueue("a"); q.enqueue("b"); q.enqueue("c"); while(! q.isempty()) { System.out.print(q.dequeue() + " "); 1

2 3. Complete method that will cause every element in the LinkedQueue<E> argument to be duplicated. If the LinkedQueue contains the objects "a", "b", "c" (with "a" at the front of the queue and "c" at the end) the LinkedQueue should be made to contain "a", "a", "b", "b", "c", "c" (with "c" still at the end). The object referenced by q could have any number of elements in it. These assertions must pass. (14pts) OurQueue<String> q = new LinkedQueue<String>(); q.add("a"); q.add("b"); q.add("c"); doublequeue(q); assertequals("a", q.remove()); assertequals("a", q.remove()); assertequals("b", q.remove()); assertequals("b", q.remove()); assertequals("c", q.remove()); assertequals("c", q.remove()); asserttrue(q.isempty()); // Complete this method public void doublequeue(ourqueue<e> q) { 4. Write the return values from each call to the method named. (8pts) mystery(0) mystery(1) mystery(2) mystery(3) mystery(4) public int mystery(int n) { if (n <= 0) return 1; else return 3 + mystery(n - 1); 5. Write the output of with arguments 1, 2, and 3 (8pts) public static void stars(int n) { if (n > 1) stars(n-1); for (int i = 0; i < n; i++) System.out.print("*"); System.out.println(); 2

3 6. Write recursive method so it displays all the numbers from the first argument to the last in ascending order (separate with a space). Use recursion, do NOT use a loop. (10pts) goingup(1, 5); // goingup(2, 7); // goingup(3, 3); // 3 Additional recursion questions that will help on the final: The 8 recursive methods from RecursionFun The JavaBats from Recursion Use the following binary tree to write out each of the three traversals indicated below. (12pts) Preorder traversal Inorder traversal Postorder traversal 8. Is this a Binary Search Tree assuming only String keys are shown? (yes or no)? (3pts) 9. Is this a Binary Search Tree assuming only String keys are shown? (yes or no)? (3pts) 10. What potential benefit is there to using a binary search tree rather than an array to store the same collection of key/value mappings of a Map? (2pts) 11. What would be required of the keys for those mappings to be stored in a binary search tree? (2pts) 3

4 12. Draw a picture of the data that results from the following code on your OrderedMap class. Be sure to have the instance variable root reference the root node with the given arrow. Show the keys only. 10pts OrderedMap<String, Integer> abst = new OrderedMap<String, Integer>(); abst.put("matrix", 5); abst.put("antz", 3); abst.put("bean", 2); abst.put("shaft", 2); abst.put("scream", 3); abst.put("titanic", 4); 13. Write the output generated by the following code (10pts) List<String> list = new ArrayList<String>(); list.add("a"); list.add("b"); list.add("c"); list.add("c"); list.remove("b"); for (int i = 0; i < list.size(); i++) { System.out.print(list.get(i) + " "); System.out.println(); System.out.println(list.contains("a")); System.out.println(list.contains("e")); 14. Complete so it returns a List that only has any and all elements that are contained in both List<Integer> arguments. These assertions must pass. (10pts) List<Integer> ts1 = new ArrayList<Integer>(); ts1.add(2); ts1.add(4); ts1.add(2); ts1.add(4); List<Integer> ts2 = new LinkedList<Integer>(); ts2.add(2); ts2.add(2); ts2.add(4); ts2.add(6); ts2.add(4); // intersection([2, 4, 2, 4], [2, 2, 4, 6, 4]) should return [2, 4] or [4, 2] List<Integer> result = intersection(ts1, ts2); (2, result.size()); (result.contains(2)); (result.contains(4)); public List<Integer> (List<Integer> s1, List<Integer> s2) { 4

5 15. Complete method in OrderedMap that returns a reference to the key with the maximum value. Write two different implementations of a private helper method. One method must have a recursive solution. The other method will have an iterative solution. These assertions should public void testmaxkey() { OrderedMap<Integer, String> inttree = new OrderedMap<Integer, String>(); assertnull(inttree.maxkey()); inttree.put(5, "A"); assertequals(new Integer(5), inttree.maxkey()); inttree.put(8, "B"); assertequals(new Integer(8), inttree.maxkey()); inttree.put(-3, "C"); assertequals(new Integer(8), inttree.maxkey()); /////////////////////////////////////////////////////////// public class OrderedMap<K extends Comparable<K>, V> { private class MapNode { private MapNode left; private MapNode right; // The references to the key/value pair of the mapping private K key; private V value; public MapNode(K thekey, V thevalue) { key = thekey; value = thevalue; left = null; right = null; // end class MapNode private MapNode root; private int size; // Create an empty tree public OrderedMap() { root = null; size = 0; a) Write the public and private methods using a solution. (10pts) 5

6 b) Write the complete public method with an solution. (10pts) 16. To the same OrderedMap class above, add method to return a List<Integer> of keys in the OrderedMap object that are greater than the argument key. The given assertions must pass. public void testkeysgreater() { OrderedMap<Integer, String> om = new OrderedMap<Integer, String>(); om.put(50, "A"); om.put(60, "B"); om.put(45, "C"); om.put(55, "D"); List<Integer> list = om. (50); asserttrue(list.contains(55)); asserttrue(list.contains(60)); assertfalse(list.contains(45)); assertfalse(list.contains(50)); // Complete this method public List<K> (K key) { 17. Is it currently snowing in Tucson, Yes or No? (6pts) 6