Question

如何获得给定数字的所有可能组合。比如我有

$arr = [ 1, 2, 3, 4]

我想在组合中没有任何重复的组合

[1] => [1]
[2] => [2]
[3] => [3]
[4] => [4]
[5] => [1, 2]
[6] => [1, 3]
[7] => [1, 4]
[8] => [2, 3]
[9] => [2, 4]
[10] => [3, 4]
[11] => [1, 2, 3]
[12] => [1, 2, 4]
[13] => [1, 3, 4]
[14] => [2, 3, 4]
[15] => [1, 2, 3, 4]

Answer 1

我希望下面的函数按照你预期的输出工作：

function get_array_combination($arr) {
    $results = array(array( ));

    foreach ($arr as $values)
        foreach ($results as $combination)
                array_push($results, array_merge($combination, array($values))); // Get new values and merge to your previous combination. And push it to your results array
    return $results;
}
$set = array('1', '2', '3', '4');
$final_array = get_array_combination($set);
echo "<pre>";
print_r(array_values(array_filter($final_array))); // Removed blank entry from array and re indexing array keys.

Answer 2

我不喜欢给定的解决方案在其循环内修改迭代变量，而且我想要一个易于理解的解决方案，以移植到其他语言。所以在这里：

<?php
function permutation(array $arr)
{
        $out=[[]];
    
        foreach($arr as $key2=> $item2){
            $copy=$out;
            foreach($copy as $k=> $v){
                array_push($copy[$k],$item2 );
            }
            array_push($out,...$copy);
            
        }
        
        return $out;
}
print_r(permutation(array(1,2,3,4)));

第二个目的是使我们更好地了解正在发生的事情。

<?php
function permutation(array $arr)
{
    
    $out=[];
    
    while(count($arr)){
        
        $out_temp=[[]];
    
        foreach($arr as $key2=> $item2){
            $copy=$out_temp;
            foreach($copy as $k=> $v){
                array_push($copy[$k],$item2 );
            }
            if($key2==0){
                unset($out_temp[0]);
            }
            array_push($out_temp,...$copy);
 
        }
        
        array_push($out,...$out_temp);
        array_shift($arr);
    }

    return $out;
}
print_r(permutation(array(1,2,3,4,5)));

获得所有可能的组合，无需重复

2 个答案: