2011年1月8日 · marcog's answer is pretty good, but I'd do it the other way around so that it's easier to add even more conditions (such as having at least one digit or whatever), and I'd use lazy quantifiers because they are cheaper for certain patterns:

2014年2月7日 · I was quite happy with your answer, because I didn't know the trick n & (n - 1).Anyway, the fastest portable way will probably be something like: test bytewise first and as soon as you get a hit, test the byte for set bits.