1. CS 180 Final Exam Review 12/(11, 12)/08

2. Announcements Final Exam  Thursday, 18 th December, 10:20 am – 12:20 pm in PHYS 112  Format 30 multiple choice questions 5 programming questions  More stress on topics (chap 11 – 15) covered after exam 2 Recursion Dynamic Data Structures Window Interfaces using Swing Applets and HTML More Swing

3. Object-Oriented Design Methods  Extract functionalities of a program which will be performed multiple times Encapsulation and Information Hiding Classes, Objects, and Polymorphism  Covered in more detail later

4. Primitives int, double, boolean, float, char, etc. These are NOT Objects  Can be compared with ==, !=, =, etc.  Wrapper classes for each primitive Why do we need them? Explicit type-casting may need to be done  byte  short  int  long  float  double

5. Objects and Classes Class variables hold references (the address) to the objects Always compare with the equals() method  Do NOT use == to compare objects Assignment operators assign references – they do not make separate copies Constructors should be used to initialize class variables Calling methods  Need object  Static methods Can use class name Cannot access non-static methods/variables inside  Pass by value always Keyword: this

6. Strings Strings are a type of object  Also should not be compared with == Important String functions  charAt  indexOf  substring  length  Concatenation operator +

7. Selection Statements Modifies the flow of control of the program if/else construct  Must have a boolean condition to check against  { } are strongly recommended, but not necessary for one line statements switch construct  Multi-way branch which makes a decision based on a char, byte, short, or int type  default case  break statement

8. Repetition statements - Loops for while do-while Beware of -  Off-by-one errors  Infinite looping  Overflow

9. Arrays Linear collection of data  Can hold primitives or class types, but must all be of the same type length tells the number of elements in the array  Member, not a method Indexed from 0 to length – 1 Multiple dimension arrays Convenient to use for-loops to iterate through members of the array Remember – arrays are reference types

10. Exceptions Use a try/catch/finally block to handle exceptions thrown by a program Use throw statement to notify caller of an error User defined exceptions must extend Exception

11. File I/O Useful classes  FileInputStream, DataInputStream, FileReader, BufferedReader, Scanner, PrintWriter, DataOutputStream, etc.  ObjectInputStream, ObjectOutputStream Used to write objects to a file Any object serialized by a stream must implement Serializable Which classes are used to write text and which are used to write binary? Always close files you open

12. Inheritance and Polymorphism Differences between abstract classes and interfaces Polymorphism can simplify code Extend specific classes from general classes Use protected keyword to protect information and methods Reuse inherited functionality from parent class. Call to base class’s constructor is always called at the first line of constructor.

13. Dynamic Data Structures Inner classes Lists  Node class  Circularly linked lists  Doubly linked lists Trie data structure ( from project 7 ) Java Collections  Vector  ArrayList  LinkedList

14. Recursion Break down a problem to one or more sub-problems Be concerned only with the current level of recursion Two necessary cases  Base case  Recursive step

15. GUI and Event-driven Programming Common classes  JFrame  JPanel  JLabel  JMenu, JMenuItem  JButton Layouts  BorderLaout  GridLayout  FlowLayout

16. Challenges

17. Types Given the following classes, which of the following declarations are valid? public interface I {…} public interface J extends I {…} public interface K {…} public abstract class A {…} public class B extends A {…} implements J, K public class C extends B {…} public class D extends A {…} implements I A a = new B(); B b = new J(); C c = new B(); B b = new C(); I i = new A(); I i = new B(); I i = new D(); K k = new C();

18. Types Given the following classes, which of the following declarations are valid? public interface I {…} public interface J extends I {…} public interface K {…} public abstract class A {…} public class B extends A {…} implements J, K public class C extends B {…} public class D extends A {…} implements I A a = new B(); B b = new J(); C c = new B(); B b = new C(); I i = new A(); I i = new B(); I i = new D(); K k = new C(); valid – B is a subclass of A invalid – cannot instantiate interfaces invalid – B is the superclass of C valid – C is a subclass of B invalid – A does not implement I valid – A implements J, and J is-a I valid – D implements I valid – C extends B which implements K

19. Recursion and Linked Lists Question- A gradebook is made out of the the following Node class. class Node { private int[] values;// these values are sorted in ascending order private Node next; public Node(int[] values, Node next) public int[] getValues() public void setValues(int[] values) public Node getNext() public void setNext(Node next) } 1) Each Node contains an array of scores that are within a certain range. 2) The gradebook contains a linked list of the Node class. For example, you may use the 1 st Node to store all the scores in the F range, the 2 nd Node to store all the scores in the D range, the 3 rd Node to store all the scores in the C range, etc.

20. Recursion and Linked Lists Question continued- Each node represents a non-overlapping group of grade values and each successive Node contains a range of group values greater than the previous Node. Write an efficient method gradeExists(int), which returns true if the grade exists within the Gradebook, and false otherwise. Since the values array is sorted, you should take advantage of this and perform a binary search for the specified value.

21. Recursion and Linked Lists Approach to the solution: The solution can be broken into two parts: 1. Finding the node which might contain the grade. Since all nodes have non-overlapping ranges (in sorted order) and the arrays in turn are sorted, we only need to compare the range of each node with the given grade. 2. Do binary search on the values in that node.

22. Recursion and Linked Lists Part 1: public boolean gradeExists(int grade) { Node temp = gradebook; while (temp != null) { int[] values = temp.getValues(); if (grade < values[0]) return false; if (grade >= values[0] && grade <= values[values.length-1]) return binarySearch(values, grade, 0, values.length-1); temp = temp.getNext(); } return false; }

23. Recursion and Linked Lists Part 2: private boolean binarySearch(int[] array, int value, int lo, int hi) { if (lo > hi) return false; if (lo == hi) return array[lo] == value; int mid = (hi + lo)/2; if (array[mid] == value) return true; else if (array[mid] > value) return binarySearch(array, value, lo, mid-1); else return binarySearch(array, value, mid+1, hi); }

24. Recursion Write a function isPalindrome(String) which takes in a String and returns true if the String is a palindrome, false otherwise. Ignore the case of the letters and skip over blanks when you make the comparison. public boolean isPalindrome(String s) { }

25. Recursion Solution: public boolean isPalindrome(String s) { if (s == null) return false; if (s.length() <= 1) return true; return ( s.charAt(0) == s.charAt(s.length()-1) && isPalindrome(s.substring(1, s.length()-1)) ); } public static void main(String[] args) {... result = isPalindrome(s.toLowerCase().replace(“ “,””)); }

26. GUI Question: Create a complete Java GUI program that allows the user to enter the text for JLabels, possibly with multiple lines, and then preview them. You should use the following design: The user can enter text in the text area. Then, when the button is pushed, this text is displayed as a label in the label preview. Title Button Text Area Label Preview Enter Text:Label Preview:

27. GUI - solution public class CH12 extends JFrame implements ActionListener { JButton theButton = new JButton( "Preview" ); JTextArea theText = new JTextArea(); JLabel theLabel = new JLabel(); public CH12() { super( "Label Previewing Program" ); setSize( 400, 200 ); //button in south JPanel aPanel = new JPanel(); aPanel.add( theButton ); this.add( "South", aPanel ); //label in north aPanel = new JPanel(); aPanel.add( new JLabel("The Label Previewer") ); this.add( "North", aPanel ); JPanel centerPanel = new JPanel( new GridLayout(1,2) ); //text area in center left aPanel = new JPanel( new BorderLayout()); aPanel.add( "North", new JLabel("Enter Text:") ); theText.setLineWrap( true ); aPanel.add( "Center", theText ); centerPanel.add( aPanel ); //preview label in center right aPanel = new JPanel( new BorderLayout()); aPanel.add( "North", new JLabel("Label Preview:") ); aPanel.add( "Center", theLabel ); centerPanel.add( aPanel ); this.add( "Center", centerPanel ); theButton.addActionListener( this ); }

28. GUI – solutioncontd. public void actionPerformed( ActionEvent e ) { theLabel.setText( theText.getText() ); } public static void main( String[] args ) { JFrame ch12Frame = new CH12(); ch12Frame.setVisible( true ); } }//end of class def

29. Arrays Write a method which takes in an array of integers and replaces the values of the array with a value c i, where c i is defined as – c i = the sum of the numbers in indices 0…i in the incoming array. public void cumulativeArray(int[] a) { }

30. Arrays Write a method which takes in an array of integers and replaces the values of the array with a value c i, where c i is defined as – c i = the sum of the numbers in indices 0…i in the incoming array. public void cumulativeArray(int[] a) { if (a.length <= 1) return; for (int k=1; k<a.length; k++) a[k] = a[k] + a[k-1]; }

31. What is the output? public class A { private int x; public static int doStuff() { x = 100; x /= 3; x++; return x; } public static void main(String[] args) { System.out.println(A.doStuff()); }

32. What is the output? public class A { private int x; public static int doStuff() { x = 100; x /= 3; x++; return x; } public static void main(String[] args) { System.out.println(A.doStuff()); } Because this method is static, it does not have access to non-static class variables. This code will not compile because x is non- static.

33. Linked Lists Given an appropriate Node class, write a method which removes every other node from the list, starting with the second. public Node removeEveryOther(Node l) { }

34. Linked Lists Given an appropriate Node class, write a method which removes every other node from the list, starting with the second. public Node removeEveryOther(Node l) { Node l1 = l; while (l1 != null && l1.next != null) { l1.next = l1.next.next; l1 = l1.next; } return l; }

35. Linked Lists Now write the same method, but recursively. public Node removeEveryOther(Node l) { }

36. Linked Lists Now write the same method, but recursively. public Node removeEveryOther(Node l) { // base case if (l == null || l.next == null) return l; // recursive case Node l1 = removeEveryOther(l.next.next); l.next = l1; return l; }