数组组合计算所有的组合形式是什么？

Question

array(2) {

  [0] => array(3) {

    [0] => string(3) "16G"

    [1] => string(3) "32G"

    [2] => string(3) "64G"

  }

  [1] => array(2) {

    [0] => string(1) "L"

    [1] => string(2) "XL"

  }

}

计算所有组合

['16G','L']

['16G','XL']

['32G','L']

['32G','XL']

['64G','L']

['64G','XL']

应该怎么组合成这样的结果

array(6) {

  [0] => array(3) {

    [0] => string(3) "16G"

    [1] => string(3) "L"

  }

  [1] => array(3) {

    [0] => string(3) "16G"

    [1] => string(3) "XL"

  }

  [2] => array(3) {

    [0] => string(3) "32G"

    [1] => string(3) "L"

  }

  [3] => array(3) {

    [0] => string(3) "32G"

    [1] => string(3) "XL"

  }

  [4] => array(3) {

    [0] => string(3) "64G"

    [1] => string(3) "L"

  }

  [5] => array(3) {

    [0] => string(3) "64G"

    [1] => string(3) "XL"

  }

 }

Answer 1

如下$input = [["16G", "32G", "64G"], ["L", "XL"]];$output = array_reduce($input, function($result, $cross_item){    if(!$result){        return array_map(function($item){            return [$item];        }, $cross_item);    }    return array_reduce($cross_item, function($acc, $target) use ($result){        return array_merge($acc, array_map(function($result_item) use ($target){            $result_item[] = $target;            return $result_item;        }, $result));    }, []);}, []);

Answer 2

矩阵乘法 ？算法导论之js实现--n×n矩阵计算

Answer 3

$arr = [["16G","32G","64G"],["L","XL"]];&nbsp; &nbsp;&nbsp;$res = [];for($i=0;$i<count($arr[0]);$i++){&nbsp; &nbsp;&nbsp;&nbsp; &nbsp; for($j=0;$j<count($arr[1]);$j++){&nbsp; &nbsp; &nbsp; &nbsp; $tem = [];&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $tem[] = $arr[0][$i];&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; $tem[] = $arr[1][$j];&nbsp; &nbsp; &nbsp; &nbsp;&nbsp;&nbsp; &nbsp; &nbsp; &nbsp; if($tem){&nbsp; &nbsp; $res[] = $tem;&nbsp; &nbsp; }&nbsp; &nbsp; }&nbsp; &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;}var_dump($res);

Answer 4

如果不需要管顺序，只需要结果，可以用位运算，来进行全组合。var rt=[];const arr= [["16G", "32G", "64G"], ["L", "XL"]];for(let i=0;i<8;i++){&nbsp; &nbsp;let a1=i & 3;//取低2位&nbsp; &nbsp;if(a1>2) continue;&nbsp; &nbsp;let a2=i>>2;&nbsp; &nbsp;let t=[arr[0][a1],arr[1][a2]];&nbsp; &nbsp;rt.push(t);}