Multi-Dimensional Arrays Rectangular & Jagged Plus: More 1D traversal

Array Recall Creating a 2D array of doubles: double[][] grid = new double[3][4]; 3 rows: indices 0 to 2, 4 columns: indices 0 to 3 grid : reference to entire array grid[0] : reference to 1 st row grid[0][2] : third element in first row all entries initialized to 0.0 1D array of Point objects: Point[] points = new Point[3]; points[0] = new Point(2, 5); points[1] = new Point(0, 0); points[2] = new Point(4, 7);

Printing 2D Arrays Arrays.toString() does not work well for 2D Produces a String representation of each element, but for a 2D array, each element is an array. Use Arrays.deepToString() instead: double[][] grid = new double[3][4]; grid[0][1] = 5; grid[2][3] = 7; System.out.println(Arrays.deepToString(grid)); Output: [[0.0, 5.0, 0.0, 0.0], [0.0, 0.0, 0.0, 0.0], [0.0, 0.0, 0.0, 7.0]]

Jagged Arrays Multi-dimensional Arrays: Number of elements in a row can vary int[][] arr = new int[3][]; arr[0] = new int[2]; // 2 columns in row 0 arr[1] = new int[4]; // 4 columns in row 1 arr[2] = new int[3]; // 3 columns in row 2 To print this array: for(int i = 0; i < arr.length; i++) { // print row i for(int j = 0; j < arr[i].length; j++) System.out.print(arr[i][j] + “\t”); System.out.println(); }

Jagged Array Example Create the following array, and the print it: Output (with tabs between values in a row): jag

Tallying Values with Arrays Problem: Write a method mostFrequentDigit that returns the digit value that occurs most frequently in a number. – Example: The number contains: one 0, two 2s, four 6es, one 7, and one 9. mostFrequentDigit( ) returns 6. – If there is a tie, return the digit with the lower value. mostFrequentDigit( ) returns 3. based on notes from Reges and Stepp, Building Java Programs

Multi-Counter Problem We could declare 10 counter variables... int counter0, counter1, counter2, counter3, counter4, counter5, counter6, counter7, counter8, counter9; But a better solution is to use an array of size 10. – The element at index i will store the counter for digit value i. – Example for : – How do we build such an array? And how does it help? based on notes from Reges and Stepp, Building Java Programs index value

An Array of Tallies // assume n = int[] counts = new int[10]; while (n > 0) { // pluck off a digit and add to proper counter int digit = n % 10; counts[digit]++; n = n / 10; } based on notes from Reges and Stepp, Building Java Programs index value

Tally Solution // Returns the digit value that occurs most frequently in n. // Breaks ties by choosing the smaller value. public static int mostFrequentDigit(int n) { int[] counts = new int[10]; while (n > 0) { int digit = n % 10; / / pluck off a digit and tally it counts[digit]++; n = n / 10; } // find the most frequently occurring digit int bestIndex = 0; for (int i = 1; i < counts.length; i++) { if (counts[i] > counts[bestIndex]) { bestIndex = i; } return bestIndex; } based on notes from Reges and Stepp, Building Java Programs

Array Histogram Question Given a file of integer exam scores, such as: Write a program that will print a histogram of stars indicating the number of students who earned each unique exam score. 85: ***** 86: ************ 87: *** 88: * 91: **** based on notes from Reges and Stepp, Building JavaPrograms

Array Histogram // Reads a file of test scores and shows a histogram of score distribution. import java.io.*; import java.util.*; public class Histogram { public static void main(String[] args) throws FileNotFoundException { Scanner input = new Scanner(new File("midterm.txt")); int[] counts = new int[101]; // counters of test scores while (input.hasNextInt()) { // read file into counts array int score = input.nextInt(); counts[score]++; // if score is 87, then counts[87]++ } for (int i = 0; i < counts.length; i++) { // print star histogram if (counts[i] > 0) { System.out.print(i + ": "); for (int j = 0; j < counts[i]; j++) { System.out.print("*"); } System.out.println(); } based on notes from Reges and Stepp, Building Java Programs

Exercises What is the output? public class Mystery { public static void main(String[] args) { int x = 1; int[] a = new int[2]; mystery2(x, a); System.out.println(x + “ “ + Arrays.toString(a)); x--; a[1] = a.length; mystery2(x, a); System.out.println(x + “ “ + Arrays.toString(a)); } public static void mystery2(int x, int[] list) { list[x]++; x++; System.out.println(x + “ “ + Arrays.toString(list)); }

Exercise Output? int[][] x = {{1, 2, 3}, {1, 2}, {1}}; int sum = 0; for(int i = 0; i < x.length; i++) { sum += x[i].length; } System.out.println(sum);