Solutions to Self-Study Exercises

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Solution 9.1

Space (in bytes) allocated for each of the following?

1. int a[] = new int[5];           // 5 * 4 = 20 bytes

2. double b[] = new double[10];    // 10 * 8 = 80 bytes

3. char c[] = new char[30];        // 30 * 2 = 60 bytes

4. String s[] = new String[10];    // 10 * 4 (reference) = 40 bytes

5. Student s[] = new Student[5];   // 5 * 4 (reference) = 20 bytes

Solution 9.2	An array containing 10 floats, 1.0 to 10.0. float farr[] = {1.0,2.0,3.0,4.0,5.0,6.0,7.0,8.0,9.0,10.0};
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Solution 9.3	Prints the first element of farr. System.out.println(farr[0]);
Solution 9.4	Assigns 100.0 to the last element in farr. farr[farr.length-1] = 100.0;
Solution 9.5	A loop to print all of the elements of farr. for (int j = 0; j < farr.length; j++) System.out.println(farr[j]);

Solution 9.6

An array containing the first 100 square roots.

double doubarr[] = new double[100]; for (int k = 0; k < doubarr.length; k++)     doubarr[k] = Math.sqrt(k+1); 

Solution 9.7

Analyzing the letter frequencies in a file.

import java.io.*; import java.util.Scanner; public static void main(String[] argv) {   Scanner fileScan;                        // To read lines of text from the file   String str;                              // To store the line of text   AnalyzeFreq af = new AnalyzeFreq();   try {                                    // Create a Scanner     File theFile = new File("freqtest.txt");     fileScan = Scanner.create(theFile);     fileScan = fileScan.useDelimiter("\r\n"); // For Windows     while (fileScan.hasNext()) { // Read and count       str = fileScan.next();       af.countLetters(str);     } // while     af.printArray();                       // Print frequencies   } catch (Exception e) {      e.printStackTrace();   } // catch() } // main() 

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Solution 9.8	Sort 24 18 90 1 0 85 34 18 with insertion sort. 24 18 90 1 0 85 34 18 // Initial 18 24 90 1 0 85 34 18 // Pass 1 18 24 90 1 0 85 34 18 // Pass 2 1 18 24 90 0 85 34 18 // Pass 3 0 1 18 24 90 85 34 18 // Pass 4 0 1 18 24 85 90 34 18 // Pass 5 0 1 18 24 34 85 90 18 // Pass 6 0 1 18 18 24 34 85 90 // Pass 7
Solution 9.9	Sort 24 18 90 1 0 85 34 18 with selection sort. 24 18 90 1 0 85 34 18 // Initial 0 18 90 1 24 85 34 18 // Pass 1 0 1 90 18 24 85 34 18 // Pass 2 0 1 18 90 24 85 34 18 // Pass 3 0 1 18 18 24 85 34 90 // Pass 4 0 1 18 18 24 85 34 90 // Pass 5 0 1 18 18 24 34 85 90 // Pass 6 0 1 18 18 24 34 85 90 // Pass 7
Solution 9.10	Code to swap two Students. Student tempStud = student1; student1 = student2; student2 = tempStud;
Solution 9.11	Implementation of the selectionSort(). public void selectionSort(int arr[]) { int smallest; // Location of smallest element for (int k = 0; k < arr.length-1; k++) { smallest = k; for (int j = k+1; j < arr.length; j++) if (arr[j] < arr[smallest]) smallest = j; if (smallest != k) { // Swap smallest and kth int temp = arr[smallest]; arr[smallest] = arr[k]; arr[k] = temp; } // if } // outer for } // selectionSort()

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Solution 9.12	After mystery(myArr,k), myArr will store 1,2,3,5,5 and k will store 3.

Solution 9.13

Binary search trace for 21 in 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28:

key  iteration   low   high   mid -------------------------------------- 21   0           0     13     6 21   1           7     13     10 21   2           7     9      8 21   3           9     9      9 21   4           10    9      failure 

Solution 9.14	A two-dimensional array with five rows of 10 integers. int int2d[][] = new int[5][10];
Solution 9.15	Prints the last integer in the third row int2d and assigns 100 to its last element. System.out.println(int2d[2][9]); int2d[4][9] = 100;
Solution 9.16	Prints all of the elements of int2d. for (int k = 0; k < int2d.length; k++) { for (int j = 0; j < int2d[k].length; j++) System.out.print( int2d[k][j] + " "); System.out.println(); // new line }

Solution 9.17	A 52 x 7 two-dimensional array of int. int sales[][] = new int[52][7]; for (int k = 0; k < sales.length; k++) for (int j= 0; j < sales[k].length; j++) sales[k][j] = 0;
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Solution 9.18	A method to calculate average number of newspapers per week. double avgWeeklySales(int arr[][]) { double total = 0; for (int k = 0; k < arr.length; k++) for (int j = 0; j < arr[k].length; j++) total += arr[k][j]; return total/52; }
Solution 9.19	A method to calculate average number of Sunday newspapers. double avgSundaySales(int arr[][]) { double total = 0; for (int k = 0; k < arr.length; k++) total += arr[k][6]; return total/52; }

Solution 9.20	A compareTo() for LetterFreq. public int compareTo(Object lf) { LetterFreq letFreq = (LetterFreq)lf; if (freq < letFreq.getFreq()) return -1; else if (freq > letFreq.getFreq()) return +1; else return 0; // The frequencies must be equal. } // compareTo()
Solution 9.21	A sort() for AnalyzeFreq. public void sort() { java.util.Arrays.sort(freqArr); } // sort()
Solution 9.22	A new AnalyzeFreq.main() that uses sort(). public static void main(String[] argv) { AnalyzeFreq af = new AnalyzeFreq(); af.countLetters("Now is the time for all good students" + " to study computer-related topics."); af.sort(); af.printArray(); } // main()