C Sc 127B Practice Test 2 Section Leader Your Name 100pts

Transcription

1 C Sc 127B Practice Test 2 Section Leader Your Name 100pts Assume we have two collection class named Stack and Queue that have the appropriate messages: Stack public boolean isempty(); public void push(e element); public E peek(); public E pop(); Queue public boolean isempty(); public void enqueue(e newel); public E peek(); public E dequeue(); 1. Write the output generated by the following code: (8pts) Stack<Character> s = new Stack<Character>(); s.push('a'); s.push('b'); s.push('c'); System.out.println(s.peek()); System.out.println(s.pop()); System.out.println(s.isEmpty()); System.out.println(s.peek()); s.push('d'); while(! s.isempty()) { haracter ch = s.peek() System.out.println(s.pop()) 2. Write the output generated by the following code. (6pts) Queue<String> q = new Queue<String>(); q.enqueue("a"); q.enqueue("b"); q.enqueue("c"); while(! q.isempty()) { System.out.println(q.isEmpty() + " " + q.dequeue()); 3. Complete method popuntil to modify the stack argument until value is found or the stack is empty. Do not allow popuntil to throw any exceptions if value is not in the stack argument. (10pts) Stack<String> s = new Stack<String>(); s.push("1"); s.push("+"); s.push("("); s.push("2"); popuntil("(", s); assertequals("+", s.pop()); assertequals("1", s.pop()); public void popuntil(string value, OurStack<String> s) { 1

2 4. Complete method doublequeue that will cause every element in the OurQueue argument to be duplicated. If the OurQueue object contains the objects "a", "b", "c", "d", "e" (with "a" at the front of the queue and "e" at the end) the LinkedQueue should be made to contain "a", "a", "b", "b", "c", "c", "d", "d", "e", "e" (with "e" still at the end). The OurQueue referenced by q could have any number of elements in it. (14pts) OurQueue<String> q = new LinkedQueue<String>(); q.add("a"); q.add("b"); doublequeue(q); assertequals("a", q.remove()); assertequals("a", q.remove()); assertfalse(q.isempty()); assertequals("b", q.remove()); assertequals("b", q.remove()); asserttrue(q.isempty()); public void doublequeue(ourqueue<string> q) { 5. Write the output generated by the method call mystery2("x", 6); (4pts) 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); 6. Implement the trib method recursively. (6pts) 2

3 7. 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. (16pts) odddownevenup(1); // 1 odddownevenup(2); // 1 2 odddownevenup(4); // odddownevenup(9); // public void odddownevenup(int n) 8. Use recursion to complete method occurrencesof that returns how often an int element occurs in an array of integers. Use recursion. Do not use a loop. (14pts) int[] a = { 1, 2, 1, 2, 4 ; assertequals( 2, occurrencesof(1, a); assertequals( 2, occurrencesof(2, a); assertequals( 1, occurrencesof(4, a); assertequals( 0, occurrencesof(999, a); public int occurrencesof(int value, int[] x) { 3

4 9. Add recursive method removeall to ArrayList<E> to remove every element that equals el. The size should decrease for each element removed. Your solution must use recursion. Existence of a loop anywhere results in a zero. public class ArrayList<E> { private Object[] data; int n = 0; public ArrayList() { data = new Object[10000]; n = 0; public int size() { return n; ArrayList<String> list = new ArrayList<String>(); list.add(0, "B"); list.add(0, "X"); list.add(0, "B"); list.add(0, "A"); assertequals("a B X B", list.tostring()); assertequals(4, list.size()); list.removeall("b"); assertequals("a X", list.tostring()); assertequals(2, list.size()); public void add(int index, E element) { // Shift array elements to the right for (int i = size(); i > index; i--) data[i] = data[i - 1]; data[index] = el; n++; // Remove every occurrence of any array element that equals el. // Use recursion, do not use a loop anywhere. public void removeall (E el) { 4

5 10) To class LinkedList class using a singly-linked structure, complete recursive method removeall(e element) so all elements in the singly linked structure that equals el are removed. The size should decrease for each element removed. (16pts) 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); n++; public int size() { return n; // Remove every occurrence of any array element that equals el. // Use recursion, do not use a loop anywhere. public void removeall (E el) { LinkedList<String> list = new LinkedList<String>(); list.addfirst("b"); list.addfirst("x"); list.addfirst("a"); list.addfirst("a"); assertequals("a A X B", list.tostring()); assertequals(4, list.size()); list.removeall("a"); assertequals("x B", list.tostring()); assertequals(2, list.size()); 5