C Sc 227 Practice Final Summer 13 Name 200 pts

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1 C Sc 227 Practice Final Summer 13 Name 200 pts 1. 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? (2pts) 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) (6pts) 2. Write Java code that will generate the following linked structure assuming you have access to class Node (4pts) 3. On the next page, write a) method reverse() so all elements are stored in reverse order. Also write b) method addlast(e) to add elements to the end of the LinkedList. The code in the box (see next page) must compile with passing assertions. You cannot pretend there are other LinkedList methods. If you need another, write it. public class LinkedList<E extends Comparable<E>> { private class Node { 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; public void addfirst(e element) { first = new Node(element, first); n++; public int size() { return n; 1

2 public String tostring() { if (first == null) return "[]"; else if (first.next == null) return "[" + first.data + "]"; else { String result = "[";d Node ref = first; while (ref.next!= null) { result += ref.data + ", "; ref = ref.next; result = result + ref.data + "]"; return result; public void reverse() { // 16 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()); public void addlast(e el) // 12 pts 2

3 4. Write the output generated by the following code. (4pts) Stack<Character> s = new Stack<Character>(); s.push('x'); s.push('y'); System.out.println(s.peek()); System.out.println(s.pop()); System.out.println(s.isEmpty()); 5. Complete method doublestack that will cause every element in the Stack argument to be duplicated. If the Stack contains the objects "a", "b", "c" (with "c" at the top of the stack) the Stack 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) Stack<String> s = new Stack<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 doublestack(stack<string> arg) { 6. Write the return values from each call to the method named mystery. (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); 3

4 7. In the box, write the output of stars with of arguments 1, 2, and 3. (6pts) public 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): 8. Write the output generated by the method call mystery2("x", 6); (8pts) public void mystery2(string s, int digit) { if(digit <= 1) System.out.println(digit); else { s = s + "<"; mystery2(s, digit - 2); System.out.println(s + digit); 9. Use recursion to complete method odddownevenup 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. (14pts) odddownevenup(1); // 1 odddownevenup(2); // 1 2 odddownevenup(3); // odddownevenup(9); // public void odddownevenup(int n) 4

5 10. Complete recursive method samestarchar to return true if for every '*' (star) in the string, if there are chars both immediately before and after the star, they are the same. Do not use any loop anywhere, use recursion. (12 pts) asserttrue(samestarchar("xy*yzz")) assertfalse(samestarchar("xy*zzz") asserttrue(samestarchar("*xa*az")) public boolean samestarchar(string str) { 11. Complete method sameends to return true if the group of len numbers at the start and end of the array are the same. For example, with {5, 6, 45, 99, 13, 5, 6, the ends are the same for n=0 and n=2, and false for n=1 and n=3. You may assume that n is in the range 0..nums.length inclusive. Do not use any loop anywhere, use recursion. (12 pts) assertfalse(sameends(new int[] { 5, 6, 45, 99, 13, 5, 6, 1)); asserttrue(sameends(new int[] { 5, 6, 45, 99, 13, 5, 6, 2)); assertfalse(sameends(new int[] { 5, 6, 45, 99, 13, 5, 6, 3)); public boolean sameends(int[] nums, int len) { 5

6 12. Use the following binary tree to write out each of the three traversals indicated below. (12pts) Preorder traversal Inorder traversal Postorder traversal 13. Is this a Binary Search Tree assuming only String keys are shown? (yes or no)? (2pts) root 14. Is this a Binary Search Tree assuming only String keys are shown? (yes or no)? (2pts) root 15. Draw a picture of the data that results from the following code on your OrderedSet class. Be sure to have the instance variable root reference the root node with an arrow. Show values. (6pts) OrderedSet<String> abst = new OrderedSet<String>(); abst.insert("matrix"); abst.insert("antz"); abst.insert("bean"); abst.insert("shaft"); abst.insert("scream"); abst.insert("titanic"); 6

7 16. In our OrderedSet<E> class, complete method nthlargest so it returns the nth largest value in the OrderedSet. For example, map.nthlargest(0) must return the key in the map that is less than all other elements and map.nthlargest(map.size()-1) must return the element that is greater than all other elements. If the int argument n is less than 0 or n >= map.size(), return null. Make your solution general enough to work for any sized OrderedSet<E> with any Comparable. You do not need to use recursion, but may. (16pts) OrderedSet<String> set = new OrderedSet<String>(); assertnull(set.nthlargest(0)); // set.size() is now 0, so nthlargest returns null set.put("m", 1); set.put("b", 2); set.put("s", 3); set.put("t", 4); assertequals("b", set.nthlargest(0)); assertequals("m", set.nthlargest(1)); assertequals("s", set.nthlargest(2)); assertequals("t", set.nthlargest(3)); assertnull(set.nthlargest(set.size())); // set.size() is too large, return null public class OrderedSet<E extends Comparable<E>> { private class TreeNode { private E data; private TreeNode left, right; public TreeNode(E el) { data = el; left = null; right = null; private TreeNode root; public OrderedSet() { root = null; public boolean insert(e element) { /*... */ public boolean contains(e element) { /*... */ public int size() { /*... */ public E nthlargest(int n) { // Complete this method 7

8 17. To our OrderedSet class add method bottomlevelfilled to the BinarySearchTree class. A bottomlevelfilled message returns true if the bottommost level of the binary search tree has a node at every possible location. The following BST would result in false since the bottom level could have four nodes instead of the three shown. An empty tree has its bottom level filled. And the following trees would result in bottomlevelfilled returning true. Assume the only method that exists in BinarySearchTree is the put method, which you won't need. (16pts) // Complete this method. You may add other methods. public boolean bottomlevelfilled() { 8

9 18. Complete the output that would be generated by this code. (6pts) 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) + " "); 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)); 19. Complete the output that would be generated by this code. (10pts) Map<Integer, String> map = new TreeMap<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(75)); System.out.println(map.get(99)); System.out.println(map.keySet()); System.out.println(map.values()); 20. Complete intersect so it returns a List that only has any and all elements that are contained in both List<Integer> arguments. (12pts) 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); assertequals(2, list.size()); asserttrue(list.contains(2)); asserttrue(list.contains(4)); public List<Integer> intersect (List<Integer> s1, List<Integer> s2) { 9