This midterm has about 4 questions more than the actual midterm of 200 pts will have

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1 C Sc 227 Practice Midterm Name 246 points This midterm has about 4 questions more than the actual midterm of 200 pts will have 1. Write a checkmark to the right of any assertion in the following test method if it would pass. Leave the line in the comment blank if the assertion would fail. public void testexpressions() { int j = 5; int k = 3; double x = -1.5; String s = "U of Arizona"; assertequals(2, j / k); assertequals(1, j % k); assertequals(-0.5, x + 1, 1e-12); assertequals('u', s.charat(1)); // a. // b. // c. // d. assertequals(5, s.indexof("arizona")); // e. assertequals(0, s.indexof("u of A")); // f. assertequals(2.0, Math.sqrt(j), 1e-12); // h. asserttrue(s.compareto("asu") > 0); // i. assertequals(6, s.compareto("u of A")); // j. assertequals(3, s.compareto("u of D")); // k. 2a. Complete the unit test for method grade (fill in the blanks) so it executes every branch of code in the nested selection. Assume the method grade will be in the same class as the method. Implement the following grading policy (10pts) Percentage Grade 90.0 >= percentage "A" 80.0 >= percentage < 90.0 "B" 70.0 >= percentage < 80.0 "C" 60.0 >= percentage < 70.0 "D" percentage < 60.0 "E" assertequals( ); assertequals( ); assertequals( ); assertequals( ); assertequals( ); 2b. Complete method grade as a method in the same class as the assertions above. (8pts) public String grade(double percent) { 1

2 3. How many times will each code fragment print "Hello"? zero and Infinite are legitimate answers (12pts, 4pts each) int j = 1; int n = 5; while(j <= n) { System.out.print("Hello"); n++; int j = 1; while(j <= 11) { System.out.print("Hello"); j = j + 3; int n = 0; for(int j = 1; j < n; j++) { System.out.print("Hello"); // Tricky Question for(int j = 1; j <= 11; j++); System.out.print("Hello"); 4. Complete method occurencesof to return how often a particular int is found in the Scanner argument's input String. These assertions that help explain the expected output must pass. (10pts) Scanner scanner = new Scanner(" "); assertequals(3, cf.occurencesof(11, scanner)); Scanner scanner2 = new Scanner(" "); assertequals(2, cf.occurencesof(6, scanner2)); // Precondition: scanner has only valid integers to scan public int occurencesof(int target, Scanner scanner) { 5. Complete method mirrorends that given a string, looks for a mirror image (backwards) string at both the beginning and end of the given string. In other words, zero or more characters at the very beginning of the given string, and at the very end of the string in reverse order (possibly overlapping). For example, "abxyzba" has the mirror end "ab". (10pts) assertequals("", cf.mirrorends("")); assertequals("", cf.mirrorends("abcde")); assertequals("a", cf.mirrorends("abca")); assertequals("aba", cf.mirrorends("aba")); public String mirrorends(string str) { 2

3 6. Complete method exists to return true if the String argument is found at least once in an array of Strings. If the string argument target is not found, return false. (10pts) String[] names = { "Li", "Devon", "Sandeep", "Chris", "Regan" ; asserttrue(af.exists("li", names, 5)); asserttrue(af.exists("sandeep", names, 5)); assertfalse(af.exists("not HERE", names, 5)); public boolean exists(string target, String[] a, int n) { 7. Complete method post4 such that when given a non-empty array of ints, post4 returns a new array containing the elements from the original array that come after the right most 4 in the original array argument. The array argument will always contain at least one 4. (14pts) assertarrayequals(new int[] {1, 2, post4(new int[]{2, 4, 1, 2)); assertarrayequals(new int[] {2, post4(new int[]{4, 1, 4, 2)); assertarrayequals(new int[] {1, 2, 3, post4(new int[]{4, 4, 1, 2, 3)); assertarrayequals(new int[] {, post4(new int[]{4)); // Precondition: The array argument is never an empty array and always contains at least one 4 public int[] post4(int[] nums) { 3

4 8. In class String227 complete the equals method to return true if two String227 objects have the same exact characters at the same indexes and their lengths are the same. Return false otherwise. You may not use any String227 methods other than the constructor, charat(int), and length(). public void testequals() { char[] a1 = { 'A', 'B', 'C' ; char[] a2 = { 'A', 'B', 'C' ; char[] a3 = { 'A', 'B', 'C', 'D' ; char[] a4 = { 'D', 'E', 'F' ; String227 str1 = new String227(a1); String227 str2 = new String227(a2); String227 str3 = new String227(a3); String227 str4 = new String227(a4); asserttrue(str1.equals(str1)); asserttrue(str1.equals(str2)); assertfalse(str1.equals(str3)); assertfalse(str1.equals(str4)); assertfalse(str1.equals(str4)); public class String227 implements Comparable<String227> { private char[] thechars; private int n; public String227(char[] a) { thechars = new char[128]; n = a.length; for (int i = 0; i < n; i++) { thechars[i] = a[i]; public char charat(int index) { return thechars[index]; public int length() { return n; public String tostring() { String result = ""; for (int i = 0; i < n; i++) result += thechars[i]; return result; public boolean equals(string227 other) 4

5 9. In the same class String227, complete method compareto to class String227 so it returns the difference of the first character found that differs, a negative int if this String227 is less than the argument, a positive int if this String227 object is greater than the argument. If all characters are exactly equal, return 0. If both String227 objects have the same exact characters up to the end of the shorter string, return the difference in lengths: "abcde".comparto("ab") returns 3 public void testcompareto() { char[] chars1 = { 'a', 'b', 'c', 'd' ; String227 str1 = new String227(chars1); char[] chars2 = { 'a', 'b', 'f', 'd' ; String227 str2 = new String227(chars2); char[] chars3 = { 'a', 'b', 'c', 'd', 'e', 'f' ; String227 str3 = new String227(chars3); assertequals(-3, str1.compareto(str2)); assertequals(+3, str2.compareto(str1)); assertequals(-2, str1.compareto(str3)); assertequals(+2, str3.compareto(str1)); assertequals(0, str3.compareto(str3)); assertequals(0, str2.compareto(str2)); assertequals(0, str1.compareto(str1)); public int compareto(string227 other) { 5

6 10. In the Matrix class, complete method putsumindiagonal that places the sum of each row on the diagonal with all other elements set to 0. It should work for any initial values set in the constructor (even if initial values are set to random integers). The output shown to the right of the method that generates that output shows the expected result public void testputsumindiagonal() { int[][] twodarray = { { 1, 2, 3, 4, { 2, 3, 4, 5, { 3, 4, 5, 6, { 4, 5, 6, 7 ; Matrix m = new Matrix(twoDarray); System.out.println(m.toString()); m.putsumindiagonal(); System.out.println(m.toString()); Output public class Matrix { private int[][] data; private int nrows; private int ncols; public Matrix(int[][] table) { nrows = table.length; ncols = table[0].length; data = new int[nrows][ncols]; for (int i = 0; i < nrows; i++) { for (int j = 0; j < ncols; j++) { data[i][j] = table[i][j]; // Makes a clone of the 2D array argument public String tostring() { String result = ""; for (int row = 0; row < nrows; row++) { for (int col = 0; col < ncols; col++) { result += " " + data[row][col]; result += "\n"; return result; public void putsumindiagonal() { 6

7 11. As part of the same Matrix class from the previous page, complete method mirrored()to change data to the mirror image of itself with the top row moved to the bottom row, the 2 nd row to the 2 nd to last row and so on. All elements in each row must also be the mirror image. For example, the 2D arrays data should be changed like these two: (16pts) data before data after data before data after public void mirrored() { // you have access to data, nrows, and ncols 12. Determine the tightest upper bound runtimes of the following loops. Express your answer in the Big-O notation we have been using in class (assume the initialization int sum = 0;). (24pts: 3pts each for order, 6pts for overall style) a. for (int j = n; j >= 0; j--) b. for (int j = 1; j <= n; j = j + 2) c. for(int k = 1; k <= n; k = 2 * k) d. for( int j = 1; j <= n; j++ ) for( int k = 1; k <= n; k++ ) for(int m = 1; m <= n; m++ ) e. int j = 1; f. for(int k = 1; k <= n; k = 2 * k) for (int j = 1; j <= 4 * n; j = j + 2) 7

8 13. Write an X in the comment to the right of all statements that compile. (3pts) Object anobject = new Integer(3); // a. Object otherobject = "a string"; // b. Integer anint = new Object(); // c. 14. Write an X in the comment to the right of all statements that compile. (3pts) Integer i1 = 3; // a. int i2 = new Integer(-1); // b. Object obj = 3; // c. 15. Write an X in the comment to the right of all statements that compile. (3pts) Object[] elements = new Object[5]; // a. elements[0] = 12; // b. elements[1] = "Second"; // c. 16. Will this code compile, yes or no? (2pts) Object obj = 3; Integer anint = obj; 17. Will this code generate a compiletime error, yes or no? (2pts) Object obj = 3; String str = (String)obj; 18. Will this code generate a runtime error (an exception) when it is run, yes or no? (2pts) Object obj = 3; String str = (String)obj 19. Write DS if it's a data structure, COLL if it's a collection class or ADT if an abstract data type (9pts) a. Bag b. singly-linked structure c. Comparable d. binary tree e. int[] f. double[][] g. Set h. LinkedSet i. PriorityList 20. What is testing for? = IllegalArgumentException.class) public void testwhat() { PriorityList<String> list1 = new ArrayPriorityList<String>(); list1.insertelementat(1, "Eric"); 21. In above, why is it possible to store an ArrayPriorityList in a variable of type PriorityList? (2pts) 8

9 22. 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) 23. Write Java code that will generate the singly-linked structure to the right assuming you have access to class Node (4pts) 24. Draw a picture of a singly-linked structure created with this code assuming you have access to class Node (4pts) Node last = new Node("A"); Node temp = new Node("B"); temp.next = last; last.next = temp; 25. 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; 9

10 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 = "[";d Node ref = first; while (ref.next!= null) { result += ref.data + ", "; ref = ref.next; result = result + ref.data + "]"; return result; 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()); public void addlast(e el) { // 10 pts 10

11 26. To the same LinkedList class on the previous page, complete method removeall(e element) as a method in LinkedList<E> so it removes all occurrences of element from any LinkedList object. Assume type E has an equals method. These assertions must pass. (16pts) LinkedList<String> list = new LinkedList<String>(); list.addfirst("b"); list.addfirst("a"); list.addfirst("b"); list.addfirst("b"); assertequals("[b, B, A, B]", list.tostring()); assertequals(3, list.size()); list.removeall("b"); assertequals("[a]", list.tostring()); assertequals(1, list.size()); public void removeall(e element) { 11