问题
Is there a regex to calculate straight poker hand?
I'm using strings to represent the sorted cards, like:
AAAAK#sssss = 4 aces and a king, all of spades.
A2345#ddddd = straight flush, all of diamonds.
In Java, I'm using these regexes:
regexPair = Pattern.compile(".*(\\w)\\1.*#.*");
regexTwoPair = Pattern.compile(".*(\\w)\\1.*(\\w)\\2.*#.*");
regexThree = Pattern.compile(".*(\\w)\\1\\1.*#.*");
regexFour = Pattern.compile(".*(\\w)\\1{3}.*#.*");
regexFullHouse = Pattern.compile("((\\w)\\2\\2(\\w)\\3|(\\w)\\4(\\w)\\5\\5)#.*");
regexFlush = Pattern.compile(".*#(\\w)\\1{4}");
How to calculate straight (sequences) values with regex?
EDIT
I open another question to solve the same problem, but using ascii value of char, to regex be short. Details here.
Thanks!
回答1:
I have to admit that regular expressions are not the first tool I would have thought of for doing this. I can pretty much guarantee that any RE capable of doing that to an unsorted hand is going to be far more hideous and far less readable than the equivalent procedural code.
Assuming the cards are sorted by face value (and they seem to be otherwise your listed regexes wouldn't work either), and you must use a regex, you could use a construct like
2345A|23456|34567|...|9TJQK|TJQKA
to detect the face value part of the hand.
In fact, from what I gather here of the "standard" hands, the following should be checked in order of decreasing priority:
Royal/straight flush: "(2345A|23456|34567|...|9TJQK|TJQKA)#(\\w)\\1{4}"
Four of a kind: ".*(\\w)\\1{3}.*#.*"
Full house: "((\\w)\\2\\2(\\w)\\3|(\\w)\\4(\\w)\\5\\5)#.*"
Flush: ".*#(\\w)\\1{4}"
Straight: "(2345A|23456|34567|...|9TJQK|TJQKA)#.*"
Three of a kind: ".*(\\w)\\1\\1.*#.*"
Two pair: ".*(\\w)\\1.*(\\w)\\2.*#.*"
One pair: ".*(\\w)\\1.*#.*"
High card: (none)
Basically, those are the same as yours except I've added the royal/straight flush and the straight. Provided you check them in order, you should get the best score from the hand. There's no regex for the high card since, at that point, it's the only score you can have.
I also changed the steel wheel (wrap-around) straights from A2345
to 2345A
since they'll be sorted that way.
回答2:
I rewrote the regex for this because I found it frustrating and confusing. Groupings make much more sense for this type of logic. The sorting is being done using a standard array sort method in javascript hence the strange order of the cards, they are in alphabetic order. I did mine in javascript but the regex could be applied to java.
hands = [
{ regex: /(2345A|23456|34567|45678|56789|6789T|789JT|89JQT|9JKQT|AJKQT)#(.)\2{4}.*/g , name: 'Straight flush' },
{ regex: /(.)\1{3}.*#.*/g , name: 'Four of a kind' },
{ regex: /((.)\2{2}(.)\3{1}#.*|(.)\4{1}(.)\5{2}#.*)/g , name: 'Full house' },
{ regex: /.*#(.)\1{4}.*/g , name: 'Flush' },
{ regex: /(2345A|23456|34567|45678|56789|6789T|789JT|89JQT|9JKQT|AJKQT)#.*/g , name: 'Straight' },
{ regex: /(.)\1{2}.*#.*/g , name: 'Three of a kind' },
{ regex: /(.)\1{1}.*(.)\2{1}.*#.*/g , name: 'Two pair' },
{ regex: /(.)\1{1}.*#.*/g , name: 'One pair' },
];
来源:https://stackoverflow.com/questions/3463964/regex-to-calculate-straight-poker-hand