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Jie
Jie

Posted on • Originally published at jma.im on

Poker StraightFlush

First of all, PHP Code Example. (PHP ≥ 7)

function isStriaghtFlush($d) {
    $b = $f = 0;
    $m = ['a'=>1,1=>10,'j'=>11,'q'=>12,'k'=>13];
    foreach ($d as $a) {
        $p = substr($a,-1);
        $b |= 1<<(($m[$a[0]])??$a[0]);
        $f = ($f===0||$f===$p) ? $p : 1;
    }
    $b = ($b << 3) | ($b >> 10);
    foreach (range(0,13) as $i) {
        if ((($b >> $i) & 31) == 31){
            return [true, $f!=1];
        }
    }
    return [false, false];
}
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Explain:

A can be straight of A, 2, 3, 4, 5 and 10, J, Q, K A

There are total 14 ranks, when getting rank n from the $d, we set 1 of the n element in bitwise map.

If we could find any 5 continuously 1 in the bitwise map, then it’s straight. also if all faces are same, then it’s straight flush.

Example, cards: 10, J, Q, K, A

read 10, bitwise map: 0,0,0,1,0,0,0,0,0,0,0,0,0,0

read J, bitwise map: 0,0,1,1,0,0,0,0,0,0,0,0,0,0

read Q, bitwise map: 0,1,1,1,0,0,0,0,0,0,0,0,0,0

read K, bitwise map: 1,1,1,1,0,0,0,0,0,0,0,0,0,0

read A, bitwise map: 1,1,1,1,0,0,0,0,0,0,0,0,1,0

shift left 3 of bitwise map to 1,1,1,1,0,0,0,0,0,0,0,0,1,0,0,0,0

shift right 10 of bitwise map to 1,1,1,1

OR this two bitwise maps and get a new map 1,1,1,1,0,0,0,0,0,0,0,0,1,1,1,1,1

then we have 5 continuously 1, it’s straight.

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