Here is some code that I came up with when trying to solve the same problem. Keep in mind that I am not a professional and do not have insane amounts of experience. It is not slow, but I'm sure it could be faster/cleaner/etc. I used it in converting voice recognized words into numbers for calculation in my own "Jarvis" a la Iron Man. It can handle numbers under 1 billion, although it could easily be expanded to include much higher magnitudes at the cost of very little time.
public static final String[] DIGITS = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
public static final String[] TENS = {null, "twenty", "thirty", "forty", "fifty", "sixty", "seventy", "eighty", "ninety"};
public static final String[] TEENS = {"ten", "eleven", "twelve", "thirteen", "fourteen", "fifteen", "sixteen", "seventeen", "eighteen", "nineteen"};
public static final String[] MAGNITUDES = {"hundred", "thousand", "million", "point"};
public static final String[] ZERO = {"zero", "oh"};
public static String replaceNumbers (String input) {
String result = "";
String[] decimal = input.split(MAGNITUDES[3]);
String[] millions = decimal[0].split(MAGNITUDES[2]);
for (int i = 0; i < millions.length; i++) {
String[] thousands = millions[i].split(MAGNITUDES[1]);
for (int j = 0; j < thousands.length; j++) {
int[] triplet = {0, 0, 0};
StringTokenizer set = new StringTokenizer(thousands[j]);
if (set.countTokens() == 1) { //If there is only one token given in triplet
String uno = set.nextToken();
triplet[0] = 0;
for (int k = 0; k < DIGITS.length; k++) {
if (uno.equals(DIGITS[k])) {
triplet[1] = 0;
triplet[2] = k + 1;
}
if (uno.equals(TENS[k])) {
triplet[1] = k + 1;
triplet[2] = 0;
}
}
}
else if (set.countTokens() == 2) { //If there are two tokens given in triplet
String uno = set.nextToken();
String dos = set.nextToken();
if (dos.equals(MAGNITUDES[0])) { //If one of the two tokens is "hundred"
for (int k = 0; k < DIGITS.length; k++) {
if (uno.equals(DIGITS[k])) {
triplet[0] = k + 1;
triplet[1] = 0;
triplet[2] = 0;
}
}
}
else {
triplet[0] = 0;
for (int k = 0; k < DIGITS.length; k++) {
if (uno.equals(TENS[k])) {
triplet[1] = k + 1;
}
if (dos.equals(DIGITS[k])) {
triplet[2] = k + 1;
}
}
}
}
else if (set.countTokens() == 3) { //If there are three tokens given in triplet
String uno = set.nextToken();
String dos = set.nextToken();
String tres = set.nextToken();
for (int k = 0; k < DIGITS.length; k++) {
if (uno.equals(DIGITS[k])) {
triplet[0] = k + 1;
}
if (tres.equals(DIGITS[k])) {
triplet[1] = 0;
triplet[2] = k + 1;
}
if (tres.equals(TENS[k])) {
triplet[1] = k + 1;
triplet[2] = 0;
}
}
}
else if (set.countTokens() == 4) { //If there are four tokens given in triplet
String uno = set.nextToken();
String dos = set.nextToken();
String tres = set.nextToken();
String cuatro = set.nextToken();
for (int k = 0; k < DIGITS.length; k++) {
if (uno.equals(DIGITS[k])) {
triplet[0] = k + 1;
}
if (cuatro.equals(DIGITS[k])) {
triplet[2] = k + 1;
}
if (tres.equals(TENS[k])) {
triplet[1] = k + 1;
}
}
}
else {
triplet[0] = 0;
triplet[1] = 0;
triplet[2] = 0;
}
result = result + Integer.toString(triplet[0]) + Integer.toString(triplet[1]) + Integer.toString(triplet[2]);
}
}
if (decimal.length > 1) { //The number is a decimal
StringTokenizer decimalDigits = new StringTokenizer(decimal[1]);
result = result + ".";
System.out.println(decimalDigits.countTokens() + " decimal digits");
while (decimalDigits.hasMoreTokens()) {
String w = decimalDigits.nextToken();
System.out.println(w);
if (w.equals(ZERO[0]) || w.equals(ZERO[1])) {
result = result + "0";
}
for (int j = 0; j < DIGITS.length; j++) {
if (w.equals(DIGITS[j])) {
result = result + Integer.toString(j + 1);
}
}
}
}
return result;
}
Input must be in grammatically correct syntax, otherwise it will have issues (create a function to remove "and"). A string input of "two hundred two million fifty three thousand point zero eight five eight oh two" returns:
two hundred two million fifty three thousand point zero eight five eight oh two
202053000.085802
It took 2 milliseconds.