C Sc 227 Practice Final

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1 C Sc 227 Practice Final Name 1. Use the Node class shown question 13 on this page 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? b) What is the value of first.next.next.data? c) What reference value equals first.next.next.next? 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) 2. Write Java code that will generate the following linked structure assuming you have access to the familiar Node class. 3. 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 end 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 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); 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()); 1

2 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() { // b) Write the addlast method here\ public void addlast(e element) { 2

3 4. Write the output generated by the following code. (8pts) 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()); 5. Complete method doublestack 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. (14pts) 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 doublestack(ourstack<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. Write the output of stars 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(); 8. Write the return values from each call to the method named mystery. mystery(25) mystery(19) mystery(9) public int mystery(int n) { if (n > 20) return 1; else if (n > 10) return 5 + mystery(n + 5); else return 3 + mystery(n + 8); 9. Write the output generated by the method call mystery2("x", 6); 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); 10. Implement the Greatest Common Divisor algorithm as public recursive method GCD. Use recursion. Do not use a loop. These assertions should pass: assertequals(3, rf.gcd(24, 9)); assertequals(99, rf.gcd(99, 99)); assertequals(10, rf.gcd(10, 30)); // Precondition: m and n are either postive of 0 (but both are not 0) public int GCD(int m, int n) 4

5 11. 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. changexy("codex") "codey" changexy("xxhixx") "yyhiyy" changexy("xhixhix") "yhiyhiy 12. 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. odddownevenup(1); // 1 odddownevenup(2); // 1 2 odddownevenup(3); // odddownevenup(4); // odddownevenup(5); // odddownevenup(9); // public void odddownevenup(int n) 5

6 13. Use the following binary tree to write out each of the three traversals indicated below. (12pts) Preorder traversal Inorder traversal Postorder traversal 14. Is this a Binary Search Tree assuming only String keys are shown? (yes or no)? (3pts) root 15. Is this a Binary Search Tree assuming only String keys are shown? (yes or no)? (3pts) root 16. Draw a picture of the data that results from the following code on the BinarySearchTree. (10pts) BinarySearchTree<String> abst = new BinarySearchTree<String>(); abst.insert("matrix"); abst.insert("antz"); abst.insert("bean"); abst.insert("shaft"); abst.insert("scream"); abst.insert("titanic"); 6

7 17. Complete intersect so it returns a LinkedSet that only has any and all elements that are contained in both Set<Integer> arguments. These assertions must pass. (16pts) public interface Set<E> { Set<Integer> ts1 = new LinkeedSet<Integer>(); ts1.add(2); ts1.add(4); ts1.add(2); ts1.add(4); Set<Integer> ts2 = new LinkedSet<Integer>(); ts2.add(2); ts2.add(2); ts2.add(4); ts2.add(6); ts2.add(4); // How many elements currently in this Set public abstract int size(); // Add element to this set only if element does // not equal another element already in this Set public abstract boolean add(e element); // Return true if element is in this Set public abstract boolean contains(e element); // Return the element at the given index so we can // access all elements in the set using a for loop public abstract E get(int index); // intersection([2, 4, 2, 4], [2, 2, 4, 6, 4]) should return [2, 4] or [4, 2] Set<Integer> result = intersection(ts1, ts2); assertequals(2, result.size()); asserttrue(result.contains(2)); asserttrue(result.contains(4)); public Set<Integer> intersect (Set<Integer> s1, Set<Integer> s2) { 18. Complete method maxkey 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()); 7

8 /////////////////////////////////////////////////////////// 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 root; private int size; // Create an empty tree public BinarySearchTree() { root = null; size = 0; a) Write the public and private methods using a recursive solution. (10pts) b) Write the complete public method with an iterative solution. (10pts) 8

9 19. To the same BinarySearchTree class above, add method Set<E> valuesgreaterthan(e key) to return a Set <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); Set<Integer> set = om.valuesgreaterthan(50); asserttrue(set.contains(55)); asserttrue(set.contains(60)); assertfalse(set.contains(45)); assertfalse(set.contains(50)); // Complete this method public Set<Integer> valuesgreaterthan(integer value) { 20. To the same BinarySearchTree class above, add method isfull if the BinarySearchTree is full. A full tree is one in which every node has exactly 2 or zero children. An empty tree is full. (16pts) public boolean isfull() { 21. Is it currently raining in Tucson (6pts) a) Yes b) No? c) Don't know for sure 9