regex - Regular expression match if there's non-alphabetical character at the end, or nothing? -

i have regular expressions match homonyms, tw?oo? match either two, to, or too. (it matches twoo, that's ok).

my question is, want regular expression match if there punctuation or other nonalphabetical character @ ends, to, or two. or ,too!. if there's nothing @ end, that's ok well.

so want match tw?oo? if there no other characters on each side, or if there non-alphabetical characters, not if there letters around: tomorrow shouldn't match.

i tried [^a-za-z]?tw?oo?[^a-za-z]? , since character classes optional ommitted.

how this, regex matches words if on own, or surrounded punctutation. (spaces aren't problem, they've been cut out)

thanks!

use word boundaries \b. match whenever word character (\w) , non-word character adjacent:

for (qw/two tomorrow/) {   "$_ ", /\b(?:two|to|too)\b/ ? "matches" : "doesn't match"; } 

output:

two matches matches tomorrow doesn't match 

edit

i changed regex /\b(?:two|to|too)\b/ per tobyink's suggestion. more readable tw?oo? , more correct tw?o+, , triggers trie optimization, transforms part of regex efficient state machine.