Faster way to find out if a number starts with 2?

java, numbers, performance

Solution

If you wanted to avoid converting it to a string, you could just keep dividing by 10 to find the most significant digit:

int getMostSignificantDigit(int x)
{
    // Need to handle Integer.MIN_VALUE "specially" as the absolute value can't
    // represented. We can hard-code the fact that it starts with 2 :)
    x = x == Integer.MIN_VALUE ? 2 : Math.abs(x);
    while (x >= 10)
    {
        x = x / 10;
    }
    return x;
}

I don't know whether this would be faster than Husman's log/pow approach.

Problem

In Java - what is the faster way to find if the given integer number is starting with the digit 2 without having to convert the number into a string? ``` String.valueOf(number).charAt(0) == '2' ```

Original source