1. Is it currently raining in Tucson (4pts) a) Yes b) No? c) Don't know d) Couldn't know (not in Tucson)
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1 1. Is it currently raining in Tucson (4pts) a) Yes b) No? c) Don't know d) Couldn't know (not in Tucson) 2. Use our familiar Node class (shown below with data and next instance variables) and this view of a linked structure to answer the questions a) through d) a) What is the value of first.data? (1pts) b) What is the value of first.next.next.data? (2pts) c) What reference value equals first.next.next.next? (2pts) d) Write the code that prints all values. Make the code general enough to work on any sized linked structure where the last node references the first (this is a circular linked list) (5pts) 3. Write Java code that will generate the following linked structure assuming (4pts) 4. On the next page, complete a) method reverse() so all elements are stored in reverse order. Also complete b) method addlast(e) to add elements to the of the LinkedList. The code in the box must compile with passing assertions, and of course both methods must work in all cases. You cannot assume there are any other LinkedList methods. If you need another method, you will have to write it. public class LinkedList<E> { private class { private E data; private Node next; public Node(E objectreference) { data = objectreference; public Node(E objectreference, Node nextreference) { data = objectreference; next = nextreference; private Node first; private int n; public LinkedList() { first = null; n = 0; 1
2 public void addfirst(e element) { first = new Node(element, first); n++; public int size() { return n; public String tostring() { if (first == null) return "[]"; else if (first.next == null) return "[" + first.data + "]"; else { String result = "["; Node ref = first; while (ref.next!= null) { result += ref.data + ", "; ref = ref.next; return result + ref.data + "]"; // a) Complete the reverse method here public void reverse() { // 12 public void testreverse() { LinkedList<String> list = new LinkedList<String>(); list.addfirst("a"); list.addfirst("b"); list.addfirst("c"); assertequals("[c, B, A]", list.tostring()); list.reverse(); assertequals("[a, B, C]", public void testaddlast() { LinkedList<String> list = new LinkedList<String>(); list.addlast("a"); list.addlast("b"); list.addlast("c"); list.addlast("d"); assertequals(4, list.size()); // Is size correct? assertequals("[a, B, C, D]", list.tostring()); // b) Write the addlast method here public void (E element) { // 12pts 2
3 5. Write the output generated by the following code. (4pts) OurStack<Character> s = new LinkedStack<Character>(); s.push('x'); s.push('y'); System.out.println(s.peek()); System.out.println(s.pop()); System.out.println(s.isEmpty()); 6. Complete method that will cause every element in the LinkedStack argument to be duplicated. If the LinkedStack contains the objects "a", "b", "c" (with "c" at the top of the stack) the LinkedStack should be made to contain "a", "a", "b", "b", "c", "c" (with the "c"s still at the top). The object referenced by s could have any number of elements in it. These assertions must pass. (12pts) OurStack<String> s = new LinkedStack<String>(); s.push("a"); s.push("b"); s.push("c"); doublestack(s); assertequals("c", s.pop()); assertequals("c", s.pop()); assertequals("b", s.pop()); assertequals("b", s.pop()); assertequals("a", s.pop()); assertequals("a", s.pop()); asserttrue(s.isempty()); // Complete this method public void (OurStack<String> arg) { 3
4 7. 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); 8. In the box to the right, write the output of with of arguments 1, 2, and 3. (5pts) 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(); stars(1) stars(2) stars(3) 9. Write the output generated by the method call ("X", 6); (7pts) public void (String s, int digit) { if(digit <= 1) System.out.println(digit); else { s = s + "<"; mystery2(s, digit - 2); System.out.println(s + digit); 10. Given a string, compute recursively (no loops) a new string where all the lowercase 'x' chars have been changed to 'y' chars. Use recursion. Do not use a loop. (10pts) changexy("codex") "codey" changexy("xxhixx") "yyhiyy" changexy("xhixhix") "yhiyhiy 4
5 11. Use recursion to complete method so it first prints all odd integers from the largest odd number <= argument down to 1. The same method call must then print all the even integers in the range of 2 through the largest even integers <= argument. The following method calls shown to the left must generate the output shown in comments to the right. Do not use any loop anywhere, use recursion. (16pts) (1); // 1 odddownevenup(2); // 1 2 odddownevenup(3); // odddownevenup(4); // odddownevenup(5); // odddownevenup(9); // Use the following binary tree to write out each of the three traversals indicated below. (12pts) Preorder traversal Inorder traversal Postorder traversal 5
6 13. Is this a Binary Search Tree assuming only String keys are shown? (yes or no)? (3pts) 14. Is this a Binary Search Tree assuming only String keys are shown? (yes or no)? (3pts) 15. Draw a picture of the data that results from the following code on your BinarySearchTree class. Be sure to have the instance variable reference the node with an arrow. Show values. (8pts) 16. Complete the output that would be generated by this code. (8pts) ArrayList<Integer> list = new ArrayList<Integer>(); list.add(12); list.add(-5); list.add(23); list.add(-5); Collections.sort(list); for (int i = 0; i < list.size(); i++) System.out.print(list.get(i) + " "); was get(0), changed to get(i) to avoid tricky question System.out.println("\n==="); System.out.println(list.contains(-5)); System.out.println(list.contains(9999)); System.out.println(Collections.min(list)); System.out.println(Collections.max(list)); 6
7 17. Complete the output that would be generated by this code. (6pts) HashMap<Integer, String> map = new HashMap<Integer, String>(); map.put(50, "M"); map.put(25, "B"); map.put(50, "C"); map.put(75, "D"); System.out.println(map.size()); System.out.println(map.get(25)); System.out.println(map.get(50)); System.out.println(map.get(75)); System.out.println(map.get(99)); 18. Complete intersect so it returns a List that only has any and all elements that are contained in both List<Integer> arguments. public void testintersect() { 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); A fewarraylist methods: // intersection([2, 4, 2, 4], [2, 2, 4, 6, 4]) should return [2, 4] or [4, 2] List<Integer> result = intersection(ts1, ts2); assertequals(2, result.size()); asserttrue(result.contains(2)); asserttrue(result.contains(4)); public List<Integer> intersect (List<Integer> s1, List<Integer> s2) { // How many elements currently in this Set public int (); // Add element to this set only if element does // not equal another element already in this Set public boolean (E element); // Return true if element is in this Set public boolean (E element); // Return the element at the given index so we can // access all elements in the set using a for loop public E (int index); 7
8 19. Complete method in BinarySearchTree 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() { BinarySearchTree<Integer > inttree = new BinarySearchTree<Integer>(); assertnull(inttree.maxkey()); inttree.insert(5); assertequals(new Integer(5), inttree.maxkey()); inttree.insert(8); assertequals(new Integer(8), inttree.maxkey()); inttree.insert(-3); assertequals(new Integer(8), inttree.maxkey()); /////////////////////////////////////////////////////////// public class BinarySearchTree<E extends Comparable<E>> { private class TreeNode { private TreeNode left; private TreeNode right; private E data; public TreeNode(K element) { key = thekey; value = thevalue; left = null; right = null; // end class TreeNode private TreeNode ; private int size; // Create an empty tree public BinarySearchTree() { = null; size = 0; a) Write the public and private methods using a solution. (8pts) 8
9 b) Write the complete public method with an solution. (8pts) 20. To the same BinarySearchTree class above, add method List to return a List<Integer> of E in the BinarySearchTree object that are greater than the argument key. The given assertions must pass. public void testkeysgreater() { BinarySearchTree<Integer> om = new BinarySearchTree<Integer>(); om.insert(50); om.insert(60); om.insert(45); om.insert(55); List<Integer> set = om. (50); asserttrue(set.contains(55)); asserttrue(set.contains(60)); assertfalse(set.contains(45)); assertfalse(set.contains(50)); // Complete this method public List<Integer> (Integer value) { 9
10 21. To the same BinarySearchTree class, add method that changes the ordering property of a BST such that an inorder traversal would visit all elements in descending order. Empty tree and trees with only one node will not change. Trees with two or more nodes will change. (16pts) \ \ M becomes M / \ G G \ \ 99 becomes 99 / \ / \ \ \ M becomes M / \ / \ G S S G / \ / \ P V V P public void reverse() { 10
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