1 / 29

Chapter 8 One-Dimensional Arrays

Chapter 8 One-Dimensional Arrays. Lecture Slides to Accompany An Introduction to Computer Science Using Java (2nd Edition) by S.N. Kamin, D. Mickunas, E. Reingold. Chapter Preview. In this chapter we will: introduce the array as a structure for storing large amounts of data

beck-russell
Download Presentation

Chapter 8 One-Dimensional Arrays

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 8One-Dimensional Arrays Lecture Slides to Accompany An Introduction to Computer Science Using Java (2nd Edition) by S.N. Kamin, D. Mickunas, E. Reingold

  2. Chapter Preview In this chapter we will: • introduce the array as a structure for storing large amounts of data • discuss common array operations • introduce algorithms for searching and sorting arrays • show how multiple images can be painted from an array to use in programming simple animations

  3. Array Declarations • Arrays contain a fixed number of variables of identical type • Array declaration and allocation are separate operations • Declaration examples: int[] counts; double[] scores; String[] studentNames;

  4. Array Allocation • Arrays are allocated using the Java new operator • The syntax is: new type[size]; • Examples: counts = new int[10]; scores = new double[15]; studentNames = new String[10];

  5. Array Organization 0 1 2 3 4 5 6 7 8 9 counts • Each box is an int variable • The numbers on top are each variable’s subscript or index • An array of size 10 has subscripts 0 to 9

  6. Array Subscripts • Arrays can contain any one type of value (either primitive values or references) • Subscripts are used to access specific array values • Examples: counts[0] // first variable in counts counts[1] // second variable in counts counts[9] // last variable in counts counts[10] // error – trying to access // variable outside counts

  7. Expressions as Subscripts • Array subscripts do not have to be constants • Array subscripts do need to be integer expressions that evaluate to valid subscript values for the current array allocation • Examples: counts[i] counts[2*i] counts[I/2]

  8. Array Initialization • Arrays can be initialized by giving a list of their elements • If your list contains n elements the subscripts will range from 0 to n – 1 • You do not need to allocate the array explicitly after it is initialized • Example: int [] primes = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29};

  9. Initializing an Array of Strings final Sring[ ] NAME = { “Sunday”, “Monday”, “Tuesday”, “Wednesday”, “Thursday”, “Friday”, “Saturday”}; // procedure that prints the day of week public void printName (int day, OutputBox out) { out.print(NAME[day – 1]); }

  10. Aliases • It is possible to have two different variables refer to the same array • When this happens these variables are called aliases • Creating aliases is not a good programming practice • Example of how it can happen: int [ ] A, B; … B = new int [10]; A = B;

  11. Loops and Array Processing • Initializes counts to 0, 10, 20, … , 90 for (int i=0; i < 10; i++) { counts[i] = i * 10; } • Prints the contents of counts using length for (int i=0; i < counts.length; i++) { out.println(counts[i]); }

  12. Extra Capacity Array • Arrays cannot grow after they have been allocated • You can allocate more space than you believe your application will need • If you guess too low, you will still run out of space • You do not need to use all the elements in an array (but total computer memory is finite)

  13. Searching • This loop terminates as soon as it finds the value 90 stored in grades bool found = false; int i = 0; while (i < size && !found) { // 90 is not in grades[0]..grades[i - 1] if (grades[i] == 90) found = true; else i++; }

  14. Processing Parallel Arrays • This loop counts the number of students whose performance improved from the first test to the second int improved = 0; for (int i = 0; i < size; i++) { if (grades1[i] < grades2[i]) improved++; }

  15. Arrays of Objects • Arrays of objects are declared in the same manner as arrays of primitive variables • Assuming that a class Student was declared elsewhere a client application could declare and allocate an array of 10 students using Student[ ] students; students = new Student[10];

  16. Passing Arrays as Arguments • When an array is passed as an argument to a method, what is passed is a pointer to the array, not a new array (arrays are passed by reference) • This means that if the method makes changes to the array, these changes are still in effect when the method returns to its caller • This is not true for primitive values which are passed by value and not by reference • Method header example: public void read(Student[ ] students) {

  17. Selection Sort • Find the smallest element among the elements A[0]..A[n-1] and call it A[min] • Swap A[0] and A[min] so A[0] contains the smallest element and A[1]..A[n-1] not sorted • Now the smallest element among the elements A[1]..A[n-1] and call it A[min] • Swap A[1] and A[min] so A[1] contains the second smallest element and A[2]..A[n-1] not sorted • Proceed similarly for A[3], A[4], and so on

  18. SelectionSort Class – part 1 public class SelectionSort { public void selectionSort (double[] A, int size) { for (int i=0; i < size; i++) { // elements in A[0]..A[i–1] are less than // elements in A[i]..A[size-1] and // A[0]..A[i-1] are sorted int min = findMinimum(A, i, size); swap(A, I, min); } }

  19. SelectionSort Class – part 2 int findMinimum (double[] A, int i, int size) { int j, min = 1; for (j= i + 1; j < size; j++) // A[min] <= all elements in A[0]..A[j–1] if (A[j] < A[min]) min =j; return min; } void swap (double[] A, int I, int j) { double temp = A[i]; A[i] = A[j]; A[j] = temp; } }

  20. Insertion Sort • Iterate over subscripts 1 to n-1 • At the ith step, shift the elements A[0]..A[i] so that this part of the array is sorted • Note: when the ith iteration begins elements A[0]..A[i – 1] are already sorted, so the only shifting is that required to insert A[i] into A[0]..A[i-1]

  21. InsertionSort Class class InsertionSort { public void InsertionSort (double[] A, int size) { int i,j; for (int i=1; i < size; i++) { double Ai = A[i]; j = j - 1; while (j >= 0 && A[j] > Ai) { A[j + 1] = A[j]; j--; } A[j + 1] = A[i]; } } }

  22. Linear Search int linearSearch (int[] A, int key) int i; for (i = 0; i < A.length; i++) { // key not in A[0]..A[i - 1] if (A[i] == key) return i; // key not in A return –1; }

  23. Searching • When searching for int values you can test for exact matches using (A[i] == key) • Comparing two double values for equality will not always give a correct result • A better comparison for doubles would be (Math.abs(A[i] – key) < epsilon) • Epsilon should be as small a value as is acceptable for your application

  24. Using Arrays in Animation Image[] mouse = new Image[NUMBER]; Int[] sleepTime = {1540, 240, 240, 240, 240, 240, 240, 240, 240, 240, 240, 240, 240, 240, 240}; public void show( ) { for (int i = 0; i < NUMBER; i++) mouse[i] = Toolkit.getDeafaultTool(). getImage(“images/T”+(i+1)+”gif); ticker = 0; do { g.drawImage(mouse[ticker}, 70, 70); Timer.pause(sleepTime{ticker]); ticker = (ticker + 1) % NUMBER; } while (true); } }

More Related