Computer Science Final Flashcards

Question

What is an object?

Answer 1

An object is an instance of a class

Answer 2

Analysis Design Coding Testing Operation Maintenance

Answer 3

In java there are numerous event sources, including JButton, JTextField, JCheckBox, etc. When the user interacts with these event sources, they fire off events. In order for these events to affect anything, we must attach event listeners to the event sources. Java then automatically attaches event handlers to the event listeners through the use of interface classes. Thus, when an event source fires off an event, the attached event listener will be triggered, which will call the event handler.

Answer 4

- Improves the reliability (robustness) of our programs - Provides useful information to client programmers (notifies them if they misuse our program)

Answer 5

- Checked - Unchecked

Answer 6

- Bubble Sort - Insertion Sort - Selection Sort

Answer 7

Checked exceptions are checked at compilation and must be caught or propagated, while unchecked exceptions are not checked at compilation and do not need to be caught.

Answer 8

1) Construction Phase - Create a heap (binary tree) - Make sure the heap meets two constraints: 1) Structural Constraint: Heap is orgranized in a top-down, left-to-right fashion 2) Value-Relationship Constraint: Heap is organized so that the value of each parent node is greater than its children's values - Start with the bottom-most, right-most parent node - Make sure its value is greater than that of its children - If its value is not greater, swap it with the child node that has the greater value and propagate it down. - Repeat for each parent node, moving in a leftward / upward motion 2) Extraction Phase 1) Extraction Step - Extract the largest value (top-most node) and put it in the output area 2) Reconstruction Step - Replace the top-most node with the bottom-most, right-most node. - Reevaluate the necessary parts of the heap - Compare the top-most node with its children. If it is smaller than its children, swap it with the larger child. - Keep comparing the node with its children until it is no longer smaller than its children or no longer has children. Repeat the extraction step until there are no more nodes remaining in the heap

Answer 9

Allows for flexibility and customization Allows Java to provide services for us

Answer 10

A sentinel-controlled loop continues as long as the given condition is true. A count-controlled loop continues for a certain number of iterations.

Answer 11

/\* Meaning of variables: from : pole from which to move disc to : pole disc is being moved to spare : pole not involved in disc transfer n : number of discs on pole/disc number \*/ public long counter = 0; public void play(int n, int from, int to, int spare) { if (n == 1) { moveOne( from, to ); } else { play( n-1, from, spare, to); moveOne( from, to); play( n-1, spare, to, from); } } private void moveOne (int from, int to) { System.out.println( from + " --\> " + to ); counter++; } public static void main(String[] args) { TOH game1 = new TOH(); game1.play(4, 1, 3, 2); System.out.println(game1.counter); }

Answer 12

Side-by-side: Different modules applied to different sub-problems Outside-to-inside: Same module applied to smaller and smaller sub-problems

Answer 13

Divide the unsorted list into 2 sub-lists recursively until each sub-list contains one element Then, merge sub-lists to produce new sorted sub-lists until there is only one list remaining

Answer 14

- Select a pivot (first of the list) - Create 2 sub-lists - All #'s less than the pivot in list 1 - All #'s greater than the pivot in list 2 - Call quick sort on list 1 and list 2 - Concatenate the result of quicksort(list1), the pivot, and quicksort(list2)

Answer 15

Basic Sort: O(n^2) Advanced Sort: O(nlogn) Basic Search: O(n) Binary Search: O(logn)

Answer 16

No (Constructor)