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Optional. Mark along the line to show your feelings Before exam: [/____________________-]. on the spectrum between / and -. After exam: [/____________________-]. UC Berkeley – Computer Science CS61B: Data Structures Midterm #1, Spring 2016 This test has 9 questions worth a total of 40 points. The exam is closed book, except that you are allowed to use a one page written cheat sheet. No calculators or other electronic devices are permitted. Give your answers and show your work in the space provided. Write the statement out below, and sign once you’re done with the exam. Write the statement out below in the blank provided and sign. You may do this before the exam begins. “I have neither given nor received any assistance in the taking of this exam.”
Signature: ___Alice Sheng__________ Name: Alice Sheng SID: 18675309 Three-letter Login ID: hug Login of Person to Left: dad Login of Person to Right: mom Exam Room: __779 Soda_________ Tips: x There may be partial credit for incomplete answers. Write as much of the solution as you can, but bear in mind that we may deduct points if your answers are much more complicated than necessary. x There are a lot of problems on this exam. Work through the ones with which you are comfortable first. Do not get overly captivated by interesting design issues or complex corner cases you’re not sure about. x Not all information provided in a problem may be useful. x Unless otherwise stated, all given code on this exam should compile. All code has been compiled and executed before printing, but in the unlikely event that we do happen to catch any bugs in the exam, we’ll announce a fix. Unless we specifically give you the option, the correct answer is not ‘does not compile.’ x The last two problems are the “hard” ones. Points Points 0 0.5 5 5. 1 3.5 6 8 2 2.5 7 0 3 5.5 8 6 4 2.5 9 6 Total 40
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0. So it begins (0.5 points). Write your name and ID on the front page. Circle the exam room. Write the IDs of your neighbors. Write the given statement. Sign when you’re done with the exam. Write your login in the corner of every page. Enjoy your free half point -.
- IntList Essentials ( 3 .5 Points). a) Recall the definition of the IntList class: public class IntList { public int val; public IntList tail; public IntList(int val, IntList tail) { this.val = val; this.tail = tail; } } What will be printed by the code below? Write your answer in the three blanks provided. public static void next1(IntList list) { list = list.tail; } public static IntList next2(IntList list) { list = list.tail; return list; } public static void next3(IntList list) { IntList temp = list; temp = temp.tail; } public static void main(String[] args) { IntList L = new IntList(1, null); L = new IntList(2, L); L = new IntList(3, L); next1(L); System.out.println(L.val); _____ 3 _______ next2(L); System.out.println(L.val); _____ 3 _______ next3(L); System.out.println(L.val); _____ 3 _______ }
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- SList Debugging and Testing (2.5 Points). Consider the SList implementation below. public class SList { public class IntNode { public int item; public IntNode next; public IntNode(int i, IntNode n) { item = i; next = n; } } private static IntNode sentinel; // static sentinel is the flaw public SList() { sentinel = new IntNode(982734, null); } public void insertFront(int x) { sentinel.next = new IntNode(x, sentinel.next); } public int getFront() { if (sentinel.next == null) { return - 1; } return sentinel.next.item; } } Write a JUnit test that fails on the code above, but would pass on a correct implementation. You may use any JUnit methods like assertEquals, assertNotEquals, assertTrue, assertFalse, etc. @Test public void test() { SList s1 = new SList(); SList s2 = new SList(); s1.insertFront(1); s2.insertFront(2); assertNotEquals(s1.getFront(), s2.getFront()); assertEquals(1, s1.getFront()); /* also fails */ }
CS61B MIDTERM, SPRING 2016
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- ArrayHorse.com ( 5 .5 Points). a) Consider the code below. Assume N is odd and positive. public static int[][] genCoolGrid(int N) { int[][] grid = new int[N][N];// Everything defaults to 0 in int arrays for (int i = 0; i <= N / 2; i += 1) { for (int j = (N / 2) - i; j <= (N / 2) + i; j += 1) { grid[i][j] = 8 ; grid[N - 1 - i][j] = 8 ; } } return grid; /* sorry, low on space, bad style, but works! */ } Show the return value of genCoolGrid(5) in the boxes below. Note: top left is grid[0][0]. grid[0][0] grid[0][4] grid[4][0] grid[4][4] Solutions that left the 0’s blank were given almost-full points. b) Suppose we define an interface for functions on integers of a single variable called IUF. An example implementation of this interface that squares integers is shown to the right. public interface IUF { public class Squarer implements IUF { int apply(int x); public int apply(int x) { return x * x; } } } Write a method mapColumn that destructively applies the given function to a single column of a 2D array. For example, if we pass in a Squarer object as f and 0 as c, we’d get back a copy of the grid with column 0 squared. public static int[][] mapColumn(IUF f, int[][] m, int c) { for(int i = 0; i < m.length; i += 1) { m[i][c] = f.apply(m[i][c]); } return m; } c) Write code such that veryCoolGrid is set equal to 5 x 5 coolGrid, but with the middle column squared. You need to instantiate a new Squarer, which is a IUF. If you used “coolGrid”, that was accepted. int[][] veryCoolGrid = mapColumn(new Squarer(), genCoolGrid(5), 2);
CS61B MIDTERM, SPRING 2016
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- EggEggeggegg (5.5 points). You start with only one variable, egg, whose contents are fully described in the following box-and-pointer diagram: a) Draw a box-and-pointer diagram if we run the following statements in the order shown: Statement A: for (int i = 0; i < 2; i += 1) {egg.tail = egg.tail.tail;} Statement B: egg.tail = new IntList(egg.tail.head - 1, egg.tail); Statement C: IntList backup = egg; Statement D: egg = egg.tail.tail; Statement E: backup.tail = egg; Descriptions of the above statements for explanation purposes: A: Deletes two nodes from the tail. B: Spawns a copy of the tail, with a value of 1 less than head. In part B, we’ll need to use this to make an IntList with a head value of 3 C: Makes a backup pointer D: Moves egg forward twice. E: Connects backup and egg To make grading easier, redraw your answer below in the box-and-pointer diagram provided. You may not need all of the boxes provided. Work outside these lines will not graded. Make sure to point the variables egg and backup at the appropriate box. Do not include boxes that cannot be reached from either egg or backup. b) Rearrange the 5 statements so that after execution is completed, there exists an IntList with exactly four elements: 1 - >2->3->4 (not necessarily pointed to by ). Write your response in the blanks below, one letter per blank. Use each letter once. C D A B E First operation Second Third Fourth Final Two alternative, correct answers: CD E AB, CDA E B
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- Array Processing ( 8 Points). You are a programmer for the people’s glorious state of Oceania. Your citizens have started trying to avoid censorship by rotating their text messages. For example, suppose a citizen wants to use the banned word “light”, for example: “nice light today”. Instead, the citizen might say “ght today nice li”. It is your job to write code to remove banned words. Throughout this problem you may assume that message and banned are of length greater than zero, and that banned.length < message.length. You may also assume that k is a valid number^1. Hint: for positive integers, a % b returns the remainder of dividing a by b, e.g. 7 % 3 is 1. (a) Fill in the matches method below, which returns true if the message starting at k matches the banned word, treating the message as circular. For example, if message is ['d', 'd', 'o', 'g', 'b', 'a'] and banned is ['b', 'a', 'd'], this method will return true for k = 4 , and false for all other k. You may not need all lines. If you don’t use a line, leave it blank. private static boolean matches(char[] message, char[] banned, int k) {
for (int i = __ 0 __; i < banned.length; i += 1) { int messageIndex = ___(k + i) % message.length____; if (message[messageIndex] != banned[i]) { ______return false_____________________; } } _____return true________________________; } (b) Complete the helper function matchStart, which returns the start index of a banned word in a message. For example, if message is ['d', 'd', 'o', 'g', 'b', 'a'] and banned is ['b', 'a', 'd'], this method should return 4 , because the match begins at position 4. If there is no match, return - 1. You can do this part even if you skipped part a! You may use the method from part a, even if you didn’t actually get it right. You may assume that banned words occur only once (if at all). public static int matchStart(char[] message, char[] banned) { for (int k = 0; k < message.length; k += 1) { if (matches(message, banned, k)) { return k; } } return - 1 ; } (^1) By valid we mean 0 ≤ k ≤ message.length – 1
UC BERKELEY Login: hug 8. Interfaces ( 6 points). You may assume that you have a correct implementation of ArrayDeque and LinkedListDeque throughout this problem, and that it implements the Deque interface. a) Consider the Deque interface from Project 1c. Add a default method reverse that reverses the Deque. public interface Deque- { void addFirst(Item x); void addLast(Item x); boolean isEmpty(); int size(); void printDeque(); Item get(int index); Item removeFirst(); Item removeLast(); default void reverse() { There were many, MANY solutions. See below for sample solutions. } } default void reverseRecursive() { // Note: if(size() != 0) is wrong if reversing two at a time if (size() > 1) { Item tempFirst = removeFirst(); reverseRecursive(); addLast(tempFirst); } } default void reverseUsingArray() { int size = size(); Item[] tmp = (Item[]) new Object[size]; for (int i = 0; i < size; i += 1) { tmp[i] = removeLast(); } for (int i = 0; i < size; i += 1) { addLast(tmp[i]); } } default void reverseUsingList(){ Deque
- helper = new LinkedListDeque
- (); // Can also use: ArrayDeque
- (); while (!isEmpty()) { helper.addLast(removeFirst()); } while (!helper.isEmpty()) { addFirst(helper.removeFirst()); } }
CS61B MIDTERM, SPRING 2016
Login: hug // "In place" solutions need two for loops: one to add, one to remove // Appending to the end. Note the for loop is different default void reverseLast2() { int size = size(); for (int i = size - 1; i >= 0; i-= 1) addLast(get(i)); for (int i = 0; i < size; i += 1) removeFirst(); } // Appending to the beginning default void reverseFirst() { int size = size(); for (int i = 0; i < size; i += 1) addFirst(get(i*2)); for (int i = 0; i < size; i += 1) removeLast(); } Designated fun zone. Draw something. Leave a scent on the paper. It is up to you. r7uvULuLUL5u:. :77LrY7L7L7u.... :.,................ :7UvuLULULUui .LLuLuLuLuLFi..iiv7uuSu2L2Jjvu77:. ..,.:::.... (^) ..... .... ...... .... ....... r2vuvuLu7jvUvju0kNjULJJXuF2P2Xu5u1YY:. rJuLuLu7uuqu5uqS0uu7uU57jvLrv7u2OE@O8Lr:i.,.. (^). ...... .... ..... .......... ,7UX2S100OjPkZqGk02qFSYSUSJXE@ZMEOFE8@M@MBM@M@Mk:.^ :1J5Y2LFPGY1S0kE2kuXF2ijLYr7YEPOq8SqFOM@GOk5:,^.. ... .... ..^.... .... ..^....^.... ..^....^............. ..^. ..^.. ... 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CS61B MIDTERM, SPRING 2016
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- Dynamic Method Selection and Inheritance ( 6 points). Modify the code below so that the max method works properly. Assume all numbers inserted into a DMSList are positive. You may not change anything in the given code. You may only fill in blanks. You may not need all blanks. public class DMSList { private IntNode sentinel; public DMSList() { sentinel = new IntNode(-1000, new LastIntNode()); } public class IntNode { public int item; /* Equivalent of first / public IntNode next; / Equivalent of rest / public IntNode(int i, IntNode h) { item = i; next = h; } public int max() { return Math.max(item, next.max()); } } public class LastIntNode extends IntNode { public LastIntNode() { super(0, null); } @Override public int max() { return 0; } } public void insertFront(int x) { sentinel.next = new IntNode(x, sentinel.next); } public int max() { return sentinel.next.max(); } public static void main(String[] args) { DMSList dmsl = new DMSList (); dmsl.insertFront(5); dmsl.insertFront(10); dmsl.insertFront(20); System.out.println(dmsl.max());/ should print 20 / } } / end of class / / end of test, boo */
