Counting Significant Figures java

Question

I am trying to make a program that counts significant figures and I am almost done there is one last thing that I am completely stuck on I can't figure out how to count the zeros between the non zero numbers here is my code:

public class Main {

    public static void main(String[] args) {
        String  Integer = "10010000000";
        String Substring = null;
        String SubString2 = null;
        if(Integer.contains(".")){
            Substring = Integer.substring(0,Integer.indexOf("."));
            System.out.println(Substring);
            SubString2 = Integer.substring(Integer.indexOf(".")+1,Integer.length());
            System.out.println(SubString2);
        }
        char [] array = Integer.toCharArray();
        char [] array1 = null;
        if(Substring !=null)
            array1 = Substring.toCharArray();
        char [] array2 = null;
        if(SubString2!=null)
            array2 = SubString2.toCharArray();
        System.out.println(amountofSigFigs(array,array1,array2,Integer));
    }
    public static int amountofSigFigs(char [] a,char [] a1,char [] a2,String Integer){
        int count = 0;
        if(a1==null && a2 == null){
            for(int i = 0;i<a.length;i++){
                if(!(a[i]=='0')){
                    count++;
                }
            }
        }
        else{
            for(int i = 0;i<a1.length;i++){
                if(!(a[i]=='0')){
                    count++;
                }
            }
            for(int i = 0;i<a2.length;i++){
                count++;
            }
        }
        return count;
    }

}

Answer 1

You shouldn't use keywords such as "Integer" for variable identifiers.

There are some lingering questions I have, but I would simply use Java's builtin split() function on strings and use regex to split the string into significant figures (ignoring ending and leading zeroes). Then return the sum of the length of the strings . Here is the code for pure integers.

String myint = "000100101111220000020000";

String[] sig_figs = myint.split("(^0+|0+$)");

int sum = 0;

for(String fig : sig_figs)
{
    sum += fig.length();
}

return sum;

If we include floats, there are other rules you would have to consider, such as zeroes after a decimal point count except if they are preceded by all zeroes (3.00 has 3, 0.03 has 1). In this case, the regex regrettably becomes much more contrived, but I would use something like "(^0+(\\.?)0*|(~\\.)0+$|\\.)" to split on. See code below (Note -- still works on integers)

String myfloat = "0.0120";

String [] sig_figs = myfloat.split("(^0+(\\.?)0*|(~\\.)0+$|\\.)");

int sum = 0;

for (String fig : sig_figs)
{
    sum += fig.length();
}

return sum;

Here's a quick look into the regex:

^0+(\\.?)0* means one or more leading zeroes (0+) followed optionally by a decimal (\\.?) and then zero or more zeroes (0*).

(~\\.)0+$ means we want to take off ending zeroes (0+$), but only if they aren't preceded by a decimal ((~\\.)).

\\. means split whatever is left by the decimal point.

Answer 2

When you find a zero, start a separate count of zeroes. Keep incrementing the zero counter instead of the main counter each time you find a zero. If you find a nonzero number, add the zeroes count value to the main counter, and set the zero counter to 0 (restart it), as well as incrementing the main counter like normal for the nonzero number. If instead, you reach the end of the array of numbers, then just ignore the zero counter.

So, you will end up adding the zero counter to the main counter only if the zeroes are not trailing zeroes, and hence, sigfigs.

Answer 3

The following code will find out the number of 0s in a string

String s = "1234320300340";
    int count = 0;
    for(int i = 0; i<s.length(); i++){

        if(s.charAt(i)== '0'){
            count++;
        }
    }
    System.out.println("total no of Zeros = "+ count);