Here’s a fast method based on Dariusz’s answer:
public static int getDigitCount(BigInteger number) {
double factor = Math.log(2) / Math.log(10);
int digitCount = (int) (factor * number.bitLength() + 1);
if (BigInteger.TEN.pow(digitCount - 1).compareTo(number) > 0) {
return digitCount - 1;
}
return digitCount;
}
The following code tests the numbers 1, 9, 10, 99, 100, 999, 1000, etc. all the way to ten-thousand digits:
public static void test() {
for (int i = 0; i < 10000; i++) {
BigInteger n = BigInteger.TEN.pow(i);
if (getDigitCount(n.subtract(BigInteger.ONE)) != i || getDigitCount(n) != i + 1) {
System.out.println("Failure: " + i);
}
}
System.out.println("Done");
}
This can check a BigInteger
with 184,948
decimal digits and more in well under a second.