Character.isDigit(string.charAt(0))
Note that this will allow any Unicode digit, not just 0-9. You might prefer:
char c = string.charAt(0);
isDigit = (c >= '0' && c <= '9');
Or the slower regex solutions:
s.substring(0, 1).matches("\d")
// or the equivalent
s.substring(0, 1).matches("[0-9]")
However, with any of these methods, you must first be sure that the string isn't empty. If it is, charAt(0)
and substring(0, 1)
will throw a StringIndexOutOfBoundsException
. startsWith
does not have this problem.
To make the entire condition one line and avoid length checks, you can alter the regexes to the following:
s.matches("\d.*")
// or the equivalent
s.matches("[0-9].*")
If the condition does not appear in a tight loop in your program, the small performance hit for using regular expressions is not likely to be noticeable.
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