Turning numbers into words

[Humphrey Bogart]

Hello. Sorry in advance for the stupid and long java question.

I want to write a java program that will turn any number of up to 9 digits in length into the correct englsh words for that number. So if the user inputs 1234, the program will respond with: one thousand two hundred and thirty four. Or if the user inputs 7231560, the program responds with: seven million, two hundred and thrity one thousand, five hundred and sixty. The caveats: the number can be of any length from 1 to 999999999, and doesn't include any commas or other delimiters.

I have successfully written a program that will do this for any number of three digits in length. I then re-wrote the program (posted below) to write a special method to do this conversion of any three digits into the proper words. I was thinking that once I had this special method, it should be easy for me to apply the three digit method to larger numbers. BUT, to apply the three digit method to a larger number, I have to split up the larger number into groups of 3, but in reverse order (so 7231560 becomes 560, 231, and 7, and not 723, 156, and zero).

SO - my question: how do I split up a number of unknown length into groups of 3 from back to front? Also - my approach of applying my three digit method to a larger number may not be the best approach to accomplish the overall task of turning a number into words. Is there a better way to do this?

Here, the program I wrote that turns any three digit number into the proper english words.

public class SayNumber {
	
	public static String Say3(String s) {
		String onePlace = "";
		String tenPlace = "";
		String hundredPlace = "";
		String wordOut = "";
		int[] tArray = new int[s.length()];
		for (int i = 0; i < s.length(); i++) 
			tArray[i] = Character.digit(s.charAt(i), 10);
 
		String[] ones = { "zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine" };
		String[] teens = { "ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen" };
		String[] tens = { null, "ten", "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety" };
		
		if (s.length() == 1){
			onePlace = (ones[tArray[0]]);
			wordOut = onePlace;
		}
		else if ( (s.length() == 2) && (tArray[1] == 0)) {
			tenPlace = (tens[tArray[0]]);
			wordOut = tenPlace;
		}
		else if ((s.length() == 2) && (tArray[0] == 1)) {
			tenPlace = (teens[tArray[1]]);
			wordOut = tenPlace;
		}
		else if ( (s.length() == 2) && (tArray[1] != 0) ) {
			tenPlace = (tens[tArray[0]]);
			onePlace = (ones[tArray[1]]);
			wordOut = (tenPlace + "-" + onePlace);
		}
		else if ((s.length() == 3) && (tArray[1] == 0) && tArray[2] == 0) {
			hundredPlace = (ones[tArray[0]]);
		 	wordOut = (hundredPlace + " hundred");
		}
		else if ((s.length() == 3) && (tArray[1] > 1) && (tArray[2] == 0)) {
			hundredPlace = (ones[tArray[0]]);
			tenPlace = (tens[tArray[1]]);
			wordOut = (hundredPlace + " hundred and " + tenPlace);
		}
		else if ((s.length() == 3) && (tArray[1] == 0)) {
			hundredPlace = (ones[tArray[0]]);
			onePlace = (ones[tArray[2]]);
			wordOut = (hundredPlace + " hundred and " + onePlace);
		}
		else if ((s.length() == 3) && (tArray[1] == 1)) {
			hundredPlace = (ones[tArray[0]]);
			tenPlace = (teens[tArray[2]]);
			wordOut = (hundredPlace + " hundred and " + tenPlace);
		}
		else if ((s.length() == 3) && (tArray[1] > 1)) {
			hundredPlace = (ones[tArray[0]]);
			tenPlace = (tens[tArray[1]]);
			onePlace = (ones[tArray[2]]);
			wordOut = (hundredPlace + " hundred and " + tenPlace + "-" + onePlace);
		}
		return wordOut; 
	}

	public static void main(String[] args) {
		String target = args[0];
					
		if (target.length() < 4) 
			System.out.print("The number is: " + Say3(target));
		else
			System.out.print("You Fool!!");
	}
}

[KevinWorkman]

Take a look at the String API for useful methods that can give you things like a String's length, a substring of a String, etc.

[Humphrey Bogart]

Thank you Kevin for your useful suggestion. I will take a look.

I think I might have solved the problem in a different fashion. It just occurred to me that if I have a number n, I can do n%1000 to give me the last three digits in their right order, no matter what the original length of n. Then (n/1000)%1000 will give me the next three digits again in proper order. I'm going to go play around with my program some more and see if I can make it work.

Thanks again.

[jim829]

Actually, n%1000 will give you any number between 0 and 999 depending on the value of n. So 10% of the possible remainders will have less than three digits.

Regards,
Jim

[JosAH]

Suppose you have a number bmmmtttuuu; if b isn't zero you pronounce it as b billiard etc. if mmm isn't zero you pronounce further as mmm million etc. the same goes for ttt thousand etc. all you have to do is write a method that can pronounce a number up to 999.

kind regards,

Jos