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9.0

Beginning Java (Histogram)

Asked by elliottbenzle in New to Java Programming, Java Programming Language, Miscellaneous Programming

Tags: Java, Histogram

I'm taking a beginner Java class, with the assignment to create a histogram program with the following output: (100 and 10 are user inputs).

How many numbers? 100
How many intervals? 10

Histogram
--------------------------------------------------------
1 ****(4)
2 ******(6)
3 ***********(11)
4 *****************(17)
5 **************************(26)
6 *************************(25)
7 *******(7)
8 ***(3)
9 (0)
10 *(1)
--------------------------------------------------------

My code is giving the following output however, can anyone help me point out what is going wrong, thanks so much.

How Many Numbers? 10
How Many Intervals? 10
Histogram
--------------------------------------------------------
1 **********(10)
2 **********(10)
3 **********(10)
4 **********(10)
5 **********(10)
6 **********(10)
7 **********(10)
8 **********(10)
9 **********(10)
10 **********(10)

For the input, 100 and 10 I get the error message:
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 10
at Lab6.main(Lab6.java:44)

Which links to this code;
intervalValue[j]++;

I'm not sure how to attach the generator (.jar) file, it is supposed to just generate the random #'s for us. Thanks again.

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             import java.util.Scanner;
 
public class Lab6 {
 
public static void main(String[] args) {
 
int numbers, intervals;
double intervalWidth;
 
double max, mins, range;
 
Scanner keyboard = new Scanner(System.in);
 
System.out.print("How Many Numbers? ");
 
numbers = keyboard.nextInt();
 
System.out.print("How Many Intervals? ");
 
intervals = keyboard.nextInt();
 
 
double [] generate = new double[numbers];
 
generate = randomGenerator(numbers);
 
 
max = maximum(generate);
 
mins = minimum(generate);
 
range = max - mins;
 
intervalWidth = range / intervals;
 
int [] intervalValue = new int[intervals];
 
for (int i=0; i < generate.length; i++) {
 
    for (int j = 0; j<generate.length; j++){
        double imin = mins+j*intervalWidth;
        double imax = max +j*(intervalWidth);
        if(generate[i] >= imin && generate[i] < imax)
            intervalValue[j]++;
        
        
    }
 
 
}
 
System.out.println("Histogram");
 
System.out.println("--------------------------------" +
 
"------------------------");
 
for (int a=0; a < intervalValue.length; a++) {
 
System.out.print(" " + (a+1) + " ");
 
for (int b=0; b < intervalValue[a]; b++) {
 
System.out.print("*");
 
}
 
System.out.println("(" + intervalValue[a] + ")");
 
}
 
}
 
private static double [] randomGenerator(int number) {
 
double [] generate;
 
generate = Generator.getData(number);
 
return generate;
 
}
 
private static double maximum(double [] a) {
 
double max = a[0];
 
for (int i = 1; i < a.length; i++) {
 
if (a[i] > max) {
 
max = a[i];
 
}
 
}
 
return max;
 
}
 
private static double minimum(double [] a) {
 
double mins = a[0];
 
for (int i = 1; i < a.length; i++) {
 
if (a[i] < mins) {
 
mins = a[i];
 
}
 
}
 
return mins;
 
}
 
}

           
         

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Beginning Java (Histogram)

Asked by elliottbenzle in New to Java Programming, Java Programming Language, Miscellaneous Programming

Tags: Java, Histogram

About this solution