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finding all series within an array

开发者 https://www.devze.com 2023-02-02 05:44 出处:网络
How do I find number of all the series (combinations of an array that have at least 3 consecutive values, like [7,8,9]) and have the longest number of values?

How do I find number of all the series (combinations of an array that have at least 3 consecutive values, like [7,8,9]) and have the longest number of values?

from [3,4,1,2,2] it would be 2 - ([1,2,3,4] twice, but ignore [1,2,3]*2 开发者_开发百科and [2,3,4]*2)

from [9,6,7,5,8] it would be 1 - ([5,6,7,8,9])

from [1,2,3,1,2] it would be 4 ([1,2,3] * 3)

Thanks


edit

the point of this is to count runs in a crib hand. It doesn't matter what order the array has the series in, as long as the series counted don't overlap all cards.


edit 1

var $cards:Array = [9, 4, 3, 2, 2];
var $ranks:Array = [];
var $c:int;
for each ($c in $cards) {
    if ($ranks[$c] == null) {
        $ranks[$c] = 1;
    }else {
        $ranks[$c] ++;
    }
}

this will create an array ($ranks) that will have these values [2:2, 3:1, 4:1, 9:1]

from this I will be able to multiply the values under 2,3 ad4 4 and multiply them by 3, so I would get 2*1*1 * 3

I'm trying to figure out now how to find the consecutive values, and ignore ones that aren't (like the 9)


You want to order your values and replace all duplicates with an array of themselves. ie.

//Order values and group matches
[3,4,1,2,2] = [1,[2,2],3,4]
[9,6,7,5,8] = [5,6,7,8,9]
[1,2,3,1,2] = [[1,1],[2,2],3]

Then you will want to find the largest consecutive sequence and remove violations.

//remove violations (6,7,8)... which I guess you also want to count separately.
[1,[2,2],3,4,6,7,8] = [1,[2,2],3,4]

Then it will be a matter of multiplying the length of all of your arrays to find your score.

Once you have cleaned your array, you can think of the formula like this.

array2 = [2,2];
array1 = [1, array2, 3, 4];
score = array1.length * array2.length = 8;

array3 = [3,3,3];
array2 = [2,2];
array1 = [1, array2, array3, 4];
score = array1.length * array2.length * array3.length = 24;

It should be pretty fun figuring out how to write this in code.


this works: it's using casalib for min/max, but there are other ways around it. this onlu finds the largest set of consecutive numbers though, as it is meant to count a crib hand which has a max of 5 cards, thus no two simultaneous series are possible (like 2,3,4 and 9,10,11)

private function countRuns($cards:Array):int {
    var $ranks:Array = [];
    var $c:int;

    for each ($c in $cards) {
        if ($ranks[$c] == null) {
            $ranks[$c] = 1;
        }else {
            $ranks[$c] ++;
        }
    }

    var $highest:int = ArrayUtil.getHighestValue($cards);
    var $lowest:int = ArrayUtil.getLowestValue($cards);
    var $seq:Array = [];
    var $longest:Array = [];
    for (var i:int = $lowest; i <= $highest; i++) {
        if ($ranks[i] != null) {
            $seq.push($ranks[i]);
            if ($seq.length > $longest.length && $seq.length > 2) {
                $longest = $seq.concat();
            }
        }else {
            $seq = [];
        }
    }

    var $total:int = $longest.length;
    for each ($c in $longest) {
        $total *= $c;
    }
    trace($total, $cards);
    return $total;
}

I found the consecutive numbers through the $seq array, by pushing values whenever the $ranks[i] has a value, if the length is greater than 3 and grater than the $longest array, copy the array over (with concat()!), if there is no value, $seq gets reset.

it's so simple once you know...


edit I noticed I had a typo in the code

if ($seq.length > $longest.length || $seq.length >= 2)

should have been

if ($seq.length > $longest.length && $seq.length > 2)

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