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CS 46B: Introduction to Data Structures June 25 Class Meeting Department of Computer Science San Jose State University Summer 2015 Instructor: Ron Mak.

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Presentation on theme: "CS 46B: Introduction to Data Structures June 25 Class Meeting Department of Computer Science San Jose State University Summer 2015 Instructor: Ron Mak."— Presentation transcript:

1 CS 46B: Introduction to Data Structures June 25 Class Meeting Department of Computer Science San Jose State University Summer 2015 Instructor: Ron Mak www.cs.sjsu.edu/~mak

2 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 1 2 public class Frequency { public static void main(String args[]) { if (args.length < 2) { System.out.println("*** Invalid arguments."); System.exit(-1); } String filePath = args[0]; String searchWord = args[1]; int count = 0; Scanner in = null; try { in = new Scanner(new File(filePath)); in.useDelimiter("[^A-Za-z]");

3 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 1, cont’d 3 try { in = new Scanner(new File(filePath)); in.useDelimiter("[^A-Za-z]"); while (in.hasNext()) { if (searchWord.equalsIgnoreCase(in.next())) ++count; } System.out.printf("The word \"%s\" appears %d times in file %s\n", searchWord, count, filePath); } catch (FileNotFoundException ex) { System.out.println("*** File not found: " + filePath); } finally { if (in != null) in.close(); } } }

4 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 2a 4 public class Name implements Comparable { private String name; private String first; private String last; public String getName() { return name; } public Name(String name) { this.name = name; String parts[] = name.split(" "); this.first = parts[0]; this.last = parts[1]; } public int compareTo(Object other) { Name otherName = (Name) other; return this.last.compareTo(otherName.last); }

5 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Questions 2b and 2c 5 public int compareTo(Object other) { Name otherName = (Name) other; return -this.last.compareTo(otherName.last); } public int compareTo(Object other) { Name otherName = (Name) other; return this.name.length() - otherName.name.length(); }

6 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 3 6 public static void main(String args[]) { Scanner in = null; try { in = new Scanner(new File(INPUT_FILE_NAME)); (new Graph(in)).printGraph(); } catch (FileNotFoundException ex) { System.out.println("*** File not found: " + INPUT_FILE_NAME); } finally { if (in != null) in.close(); }

7 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 3, cont’d 7 private void printGraph() { System.out.printf("%-12s %-5s\n\n", "LANGUAGE", "SHARE"); in.nextLine(); while (in.hasNextLine()) { Scanner line = new Scanner(in.nextLine()); line.useDelimiter("[,%]"); line.next(); String language = line.next(); float share = line.nextFloat(); long count = Math.round(share); System.out.printf("%-12s %5.2f% ", language, share); for (int i = 1; i <= count; i++) System.out.print("*"); System.out.println(); line.close(); } }

8 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 4a  Because class Bird implements interface Vocal, it must implement method vocalize(). 8 public class Bird extends Animal implements Vocal { public void move() { System.out.println("Flap, flap!"); } public interface Vocal { void vocalize(); }

9 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 4b 9 public class Mammal extends Animal implements Vocal { public void vocalize() { System.out.println("Grrr!"); } public class Dog extends Mammal { public void vocalize() { super.vocalize(); System.out.println("Arf!"); } }

10 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 4c 10 Mammal m = new Dog(); m.move(); Vocal v = new Dog(); v.vocalize(); try { v.move(); } catch (ClassCastException ex) { System.out.println("Can't move!"); }  You cannot call move() on variable v since the type of v is Vocal.  Try-catch is for catching runtime errors and doesn’t prevent compile-time errors.

11 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 5a  Possible classes (look for nouns): University Department Klass Classroom Calendar TimeOfDay Student WaitingList 11 Why the funny spelling?

12 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 5a, cont’d  Department Responsibilities  maintain list of classes  schedule classes  open classes for registration by students Collaborators  Klass  Classroom  Calendar  TimeOfDay  Students 12

13 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 5a, cont’d  Klass Responsibilities  maintain list of students  maintain waiting list Collaborators  Student  WaitingList 13

14 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 5a, cont’d  Student Responsibilities  register for classes Collaborators  Klass 14

15 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 5a, cont’d  WaitingList Responsibilities  maintain list of students Collaborators  Student 15

16 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Midterm Solution: Question 5b  Department aggregates Klass one department has many classes  Klass aggregates WaitingList each class has one waiting list  WaitingList aggregates Student one waiting list has zero, one, or more students 16

17 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Quizzes  Do quizzes 7, 8, and 9 by next 9:00 AM next Tuesday, June 30.  These cover Chapter 13, sections 13.1 – 13.4 and Worked Example 13.1 17

18 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Break 18

19 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Recursion  Recursion requires a whole new way of thinking.  Recursion is a required skill for all programmers. 19 Oh, no! Not recursion!

20 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak How to Think Recursively  Does this problem contain a simpler but similar case of the problem?  Can I solve the overall problem if I can solve the simpler case?  Is there a simplest case that has an immediate and obvious solution? This is called the base case. 20

21 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Factorials: The Classic Recursion Problem  5! = 5 x 4 x 3 x 2 x 1 = 5 x 4!  Therefore, we can solve 5! if we can solve 4! 4! is a simpler but similar case of the problem.  We can solve 4! = 4 x 3! if we can solve 3!  We can solve 3! = 3 x 2! if we can solve 2!  We can solve 2! = 2 x 1! if we can solve 1!  But by definition, 1! = 1 21

22 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Factorials, cont’d  But by definition, 1! = 1 That’s the simplest case (base case) with an immediate and obvious solution.  Therefore, 2! = 2 x 1! = 2 x 1 = 2  Therefore, 3! = 3 x 2! = 3 x 2 = 6  Therefore, 4! = 4 x 3! = 4 x 6 = 24  Therefore, 5! = 5 x 4! = 5 x 24 = 120 22

23 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Factorials, cont’d  Solve n ! recursively:  What’s the base case? 1! = 1  What’s the simpler but similar case? ( n -1)! Note that n -1 is closer to the base case of 1. 23 private int fact(int n) { if (n <= 1) return 1; else return n*fact(n-1); } Factorial.java

24 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Recursive Multiplication  Solve i x j recursively.  Base case: i equals 0: product = 0 i equals 1: product = j  Simpler but similar case: If we can solve the problem for i -1 (which is closer to 0 and 1), then i x j is [( i -1) x j ] + j 24

25 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Recursive Multiplication, cont’d 25 private long multiply(int i, int j) { switch (i) { case 0: return 0; case 1: return j; default: return j + multiply(i-1, j); } Multiplier.java

26 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Iterative Fibonacci  Fibonacci sequence: 1 1 2 3 5 8 13 21 34 55 f n = f n-2 + f n-1 f 1 = 1 f 2 = 1  An iterative solution: 26 private long fibonacci(int n) { if (n <= 2) return 1; else { long older = 1; long old = 1; long next = 1; for (int i = 3; i <= n; i++) { next = older + old; older = old; old = next; } return next; } Fibonacci.java

27 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Recursive Fibonacci  According to the definition:  f n = f n-2 + f n-1 f 1 = 1 f 2 = 1 27 private long fibonacci(int n) { if (n <= 2) return 1; else return fibonacci(n-2) + fibonacci(n-1); } FibonacciRecursive.java

28 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Recursive Fibonacci, cont’d  Why does the recursive solution take a long time when n is large?  Let’s trace the recursive calls: 28 private long fibonacci(int n) { System.out.printf("Called fibonaaci(%d)\n", n); long f; if (n <= 2) f = 1; else f = fibonacci(n-2) + fibonacci(n-1); System.out.printf("Returning fibonacci(%d) = %d\n", n, f); return f; } FibonacciTrace.java

29 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Recursive Fibonacci, cont’d 29

30 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Member of  Given a list of n integers, is x in the list?  Base case The list is empty: x is not in the list.  Simpler but similar case: Either x is equal to the first element in the list, or x is in the rest of the list. The rest of the list is one shorter, so it’s closer to the base case. 30

31 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Member of, cont’d 31 private boolean memberOf(int x, ArrayList list) { if (list.size() == 0) return false; else { int first = list.get(0); list.remove(0); return (x == first) || memberOf(x, list); } Unfortunately, this version of memberOf() destroys its list parameter. Member.java

32 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Member of, cont’d 32 private boolean memberOf2(int x, ArrayList list) { if (list.size() == 0) return false; else { int first = list.get(0); list.remove(0); return (x == first) || memberOf2(x, list); } private boolean memberOf(int x, ArrayList list) { ArrayList temp = (ArrayList ) list.clone(); return memberOf2(x, temp); } This version doesn’t harm its list parameter.

33 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Unique  Given a list of n integers in a list, remove all the duplicate values so that what remains is a list of unique values.  Base case The list is empty or it contains only one value: Just return the list (it’s empty or it has a single unique value).  Simpler but similar case: Take out the first value. Make the rest of the list unique. Then if the value we took out is not in the rest of the list, put it back. Otherwise, leave it out. 33

34 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Unique, cont’d 34 private ArrayList unique(ArrayList list) { if (list.size() <= 1) return list; else { int first = list.get(0); list.remove(0); ArrayList ulist = unique(list); // rest of list if (memberOf(first, ulist)) return ulist; else { ulist.add(0, first); // put back the first element return ulist; } Unique.java

35 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Reverse  Reverse the values of a list of n integers.  Base case The list is empty or it contains only one value: Just return the list.  Simpler but similar case: Take out the first value of the list. Reverse the rest of the list. Append the removed value to the end of the reversed rest of the list. 35

36 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Reverse, cont’d 36 private ArrayList reverse(ArrayList list) { if (list.size() <= 1) return list; else { int first = list.get(0); list.remove(0); // remove first element reverse(list).add(first); // append it to the end return list; } Reverse.java

37 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak 37 Towers of Hanoi  Goal: Move the stack of disks from the source pin to the destination pin. You can move only one disk at a time. You cannot put a larger disk on top of a smaller disk. Use the third pin for temporary disk storage.

38 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak 38 Towers of Hanoi, cont’d  Label the pins A, B, and C. A: source B: temporary C: destination  Base case: n = 1 disk Move disk from A to C (source  destination)  Simpler but similar case: n -1 disks Solve for n -1 disks (source  temp) Move disk from A to C (source  destination) Solve for n -1 disks (temp  destination

39 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak 39 Towers of Hanoi, cont’d private static final char A = 'A'; // initial source private static final char B = 'B'; // initial temp private static final char C = 'C'; // initial destination private static int count = 0; private static void move(char from, char to) { System.out.printf("%2d: Move disk from %c to %c.\n", ++count, from, to); } public static void main(String args[]) { int n = 6; System.out.printf("Solve for %d disks:\n\n", n); solve(n, A, B, C); } Hanoi1.java

40 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak 40 Towers of Hanoi, cont’d  Solve n disks (source = A, destination = C) Solve for n -1 disks (source  temp) Move disk from A to C (source  destination) Solve for n -1 disks (temp  destination) private static void solve(int n, char source, char temp, char destination) { if (n > 0) { solve(n-1, source, destination, temp); move(source, destination); solve(n-1, temp, source, destination); } Hanoi.java

41 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Homework 5: Write Recursive Methods  Method read() reads and prints a text file line by line. Its parameter is a text Scanner object. Input will be the text file GettysburgAddress.txt. Codecheck URL: http://codecheck.it/codecheck/files/15062509419psp 5nux3kfu2ypvk7t9da0fw http://codecheck.it/codecheck/files/15062509419psp 5nux3kfu2ypvk7t9da0fw 41 Note: DNS problems with http://codecheck.it For the next 48 hours, replace with http://130.211.187.232 For example: http://130.211.187.232/codecheck/files/15062509419psp5nux3kfu2ypvk7t9da0fwhttp://codecheck.ithttp://130.211.187.232 /codecheck/files/15062509419psp5nux3kfu2ypvk7t9da0fw

42 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Homework 5, cont’d  Method allSame() has a string parameter and returns true if all the characters of the string are the same, and false otherwise. Codecheck URL: http://codecheck.it/codecheck/files/15062509333k18 e6dph2n9pejrc4o49s8ax http://codecheck.it/codecheck/files/15062509333k18 e6dph2n9pejrc4o49s8ax 42

43 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Homework 5, cont’d  Method count() has two parameters, a character and a string. It returns the number of occurrences the character is in the string (case sensitive comparisons). Codecheck URL: http://codecheck.it/codecheck/files/150625093862on gmlhoag9dokoccytmfrhy http://codecheck.it/codecheck/files/150625093862on gmlhoag9dokoccytmfrhy 43

44 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Homework 5, cont’d  Method append() has two parameters that are array lists of integers. It returns an array list that is the second array list appended to the end of the first array list. Codecheck URL: http://codecheck.it/codecheck/files/1506250936deve ch3xgnds2e8b833tkb4qz http://codecheck.it/codecheck/files/1506250936deve ch3xgnds2e8b833tkb4qz 44

45 Computer Science Dept. Summer 2015: June 25 CS 46B: Introduction to Data Structures © R. Mak Homework 5, cont’d  All your methods must be recursive. You may be surprised by how short they are.  Canvas: Homework 5 Final  Due: Monday, June 29 at 11:59 PM 45

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