Computer Science Canada

Determining the amount of digits, and storing the values in array
Author:  CooKieLord [ Fri Oct 12, 2007 7:08 pm ]
Post subject:  Determining the amount of digits, and storing the values in array
Situation: I want to break down an integer number into a bunch of digits, and store the value in an array.
E.g. The user enters 1 337.

The program will break it down to 1, 3, 3, 7 and store each digit in an array such that:
integerArray[0] = 1
integerArray[1] = 3
integerArray[2] = 3
integerArray[3] = 7

I understand that I can use a combination of divisions and modulo in order to determine this, but the problem I have is that I need to determine the number of digits. This is what I have, but there's are logic errors and I can't figure them out.

Java:

public class test {
   public static void main  (String[] args) {
      int[] array;  //Array to store broken down number
      int bigNumber; //User-inout
      int digit = 1; //Number of digits
      int i =0, digitMod, expo; //dunno? O_O i is for array index, digitMod and expo are test variables
      bigNumber = ITI1120.readInt(); //ITI1120 is the class that they gave me to read numbers
     
      //Block 1 - Determine the amount of digits (Need help here)
      for (int d = 0; bigNumber % digit != digit; digit += 10) {
         digit = d; }

         array = new int[digit];

      //Block 2 - Breaking down the digits
      i = digit -1;
      expo = 1;
      while (i >= 0) {
         digitMod = Math.pow (10, expo);
         array[i] = (bigNumber / digitMod) % 10;
         expo++;
         i--;
//         digitMod += 10;
};
      //Block 3 - Printing the results
      for (int x = 0; x < digit; x++)
         System.out.print(array[x] + "\n");
     
   } // end main
}//end class
Author:  HeavenAgain [ Fri Oct 12, 2007 7:14 pm ]
Post subject:  RE:Determining the amount of digits, and storing the values in array
try making the number into a string first, then its automatically in an array
just have to convert
Author:  Euphoracle [ Fri Oct 12, 2007 7:19 pm ]
Post subject:  RE:Determining the amount of digits, and storing the values in array
Well, you could do it that way, or you could do it the easy way and use a string.

Java:

public clas test
{
        public static void main(String[] args)
        {
                int[] digits;
                int input = 0;
                int digitCount = 0;
                String quickCount = ""; // Hehehehe
                input = ITI1120.readInt(); // I would prefer to use a long, but meh
                quickCount = input + "";
                digitCount = quickCount.length(); // Oh, how I love doing things the easy way some times
                digits = new int[digitCount];
               
                for(int i = 0; i < digitCount; i++)
                {
                        digits[i] = Integer.parseInt(quickCount.charAt(i)); // :p
                }
               
                for(int i = 0; i < digitCount; i++)
                        System.out.println(digits[i]);
        }
}


Of course, if you're not allowed to do that, this won't help much Neutral Sorry for any syntax errors if any, I only have access to notepad at the moment and no JDK.
Author:  richcash [ Fri Oct 12, 2007 7:44 pm ]
Post subject:  Re: Determining the amount of digits, and storing the values in array
If you take the log (base 10) of a number and round down, the result will be 1 less than the number of digits in that number.

Remember, Math.log() in java.lang.Math is the natural log, so you have to use the base changing formula :
code:
Math.log(num)/Math.log(10)
will give you the log (base 10) of a number.

Do you know how to get the number of digits now? (Note, you'll have to make adjustments for 0 and negative numbers if you choose to include them).

As you said, you can use division and modulus once you know the number of digits.
Author:  richcash [ Sat Oct 13, 2007 12:08 pm ]
Post subject:  Re: Determining the amount of digits, and storing the values in array
In case you're having trouble :
Java:
int num = 9846711;
int numOfDigits = (int) (Math.log (num) / Math.log(10)) + 1;


Post if you need help with storing the individual digits.
Author:  wtd [ Sat Oct 13, 2007 4:05 pm ]
Post subject:  RE:Determining the amount of digits, and storing the values in array
code:
class MySpecialInteger {
    private int value;

    public MySpecialInteger() {
        this(0);
    }

    public MySpecialInteger(int value) {
        this.value = value;
    }

    public int intValue() {
        return value;
    }

    public int[] digits() {
        Integer[] temp = digitsHelper(value, new java.util.ArrayDeque<Integer>());
        int index = 0;
        int[] digits = new int[temp.length];
       
        for (Integer v : temp) {
                        digits[index++] = v;
        }
       
        return digits;
    }

    private static Integer[] digitsHelper(
      int value,
      java.util.ArrayDeque<Integer> accumulator) {
        if (value > 0) {
            accumulator.addFirst(value % 10);

            return digitsHelper(value / 10, accumulator);
        }
        else {
            return (Integer[])accumulator.toArray();
        }
    }
}


Recursion FTW!
Author:  richcash [ Sat Oct 13, 2007 5:21 pm ]
Post subject:  Re: Determining the amount of digits, and storing the values in array
I would rather write your digits() method like this.
Java:
public int[] digits() {
        int[] digits = new int[(int) (Math.log(value) / Math.log(10)) + 1];
        for (int i = 0; i < digits.length; i ++) {
                digits[i] = (int) (value / Math.pow (10, i)) % 10;
        }
        return digits;
}


But I guess thinking recursively about it is cool too.
Author:  wtd [ Sat Oct 13, 2007 6:03 pm ]
Post subject:  RE:Determining the amount of digits, and storing the values in array
You will also notice that my digitsHelper method is tail-recursive.
Author:  CooKieLord [ Tue Oct 16, 2007 2:18 pm ]
Post subject:  RE:Determining the amount of digits, and storing the values in array
Thank for all the help guys, and sorry I'm a little late in replying.

I'm mainly looking at the way richcash showed me. It's simpler for me, because I don't quite understand how to get recursion to work yet.

Now, the only problem I have is storing the broken down values in the array. With the way you're shown to me, it would place the units in [0], the 10s in [1], the 100s in [2], etc. I need the highest digit, let's say 1000s in [0], the 100s in [1], the 10s in [2] and the 1s in [3]

(I'm talking about array subscripts right? or is it called index?)

Don't tell me how to do it. I'll edit this post if I can't figure it out. I just wanted to let you guys know that I read the posts. Thanks again!
Author:  Euphoracle [ Tue Oct 16, 2007 2:26 pm ]
Post subject:  RE:Determining the amount of digits, and storing the values in array
Reverse the array:

Java:
int[] reverse(int[] source)
{
        int[] reversed = new int[source.length];
        // loop through, replace source into reversed backwards
       // code removed, sorry, didn't read that last bit
        return reversed;
}
Author:  CooKieLord [ Tue Oct 16, 2007 2:36 pm ]
Post subject:  RE:Determining the amount of digits, and storing the values in array
Well this is what I did

Java:

public class test {
   public static void main  (String[] args) {
      int[] array;
      int bigNumber;
      double digit;
      int digint;
      bigNumber = ITI1120.readInt();

      digit = Math.floor(Math.log(bigNumber) / Math.log(10)) + 1;

      array = new int[(int) digit];
      int placement = (int) array.length;

      for (int index=0; index < array.length; index ++) {
         array[index] = (int) ((bigNumber / Math.pow (10, (placement-1))) % 10);
         placement--; };

      for (int x = 0; x < array.length; x++)
         System.out.print(array[x] + "\n");

   } // end main
}//end class


I had two variables, one for the array and the other for the modulo. Is that what you meant by "reversing"?
Author:  Euphoracle [ Tue Oct 16, 2007 9:34 pm ]
Post subject:  RE:Determining the amount of digits, and storing the values in array
Your method does work, yes.

:


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