php - 在 PHP 中查找包含二维数组中每一行的一个值的每个元素组合

Question

我在我的项目中使用了更大的数组，但为了简化，假设我们有这个 3x3 数组:

$a = Array( Array(1,2,3),
            Array(4,5,6),
            Array(7,8,9) );

我想找到每行包含一个值的所有总和组合，即 1+4+7、1+4+8、1+4+9、1+5+7、1+5+8 , 1+5+9, 1+6+7, 1+6+8, 1+6+9, 2+4+7, 2+4+8, 2+4+9, 2+5+7, . ..

希望模式是显而易见的。我首先尝试了一个嵌套循环(列，然后是行)，但没有提供所有组合。经过大量搜索后，我确信解决方案需要递归，但每次尝试为此编写递归函数时，我都会感到困惑。

虽然工作代码将非常受欢迎，但对我来说可能更重要的是理解问题和解决方案。

Answer 1

给定数组中未知数量的行和列，获得所需结果的最佳方法是使用递归(否则你将不得不编写任意数量的嵌套循环)。此函数对数组中的每一行进行递归并返回两个结果之一:

1. 如果我们位于数组的最后一行，则返回该行；
2. 否则，返回当前行与数组余额结果的叉积(使用两个嵌套的foreach 循环)。我们使用 array_shift获取当前行，并将其从数组中删除。

代码如下:

function find_paths($array) {
    if (count($array) == 1) return $array[0];
    $output = array();
    foreach (array_shift($array) as $v1) {
        foreach (find_paths($array) as $v2) {
            $output[] = array_merge(array($v1), is_array($v2) ? $v2 : array($v2));
        }
    }
    return $output;
}

使用您的示例数据:

$a = Array( Array(1,2,3),
            Array(4,5,6),
            Array(7,8,9) );
$combinations = find_paths($a);

该函数执行这一系列操作:

1. 将第 1 行 (1, 2, 3) 与数组 ((4, 5, 6), (7 , 8, 9));
2. 第一次递归:将第 2 行 (4, 5, 6) 与函数输出的叉积作为数组 ( (7, 8, 9));
3. 第二次递归:此时我们在数组(7, 8, 9)中只剩下一行，所以我们返回它；
4. 第一次递归:计算 (4, 5, 6) 与 (7, 8, 9) 的叉积 = ((4, 7), (4, 8), (4, 9), (5, 7), (5, 8), (5, 9), (6, 7), (6, 8), (6, 9)) 并返回；
5. 将 (1, 2, 3) 与 ((4, 7), (4, 8), (4, 9), (5, 7) , (5, 8), (5, 9), (6, 7), (6, 8), (6, 9)) 并返回。

给出输出:

Array (
    [0] => Array ( [0] => 1 [1] => 4 [2] => 7 )
    [1] => Array ( [0] => 1 [1] => 4 [2] => 8 )
    [2] => Array ( [0] => 1 [1] => 4 [2] => 9 )
    [3] => Array ( [0] => 1 [1] => 5 [2] => 7 )
    [4] => Array ( [0] => 1 [1] => 5 [2] => 8 )
    [5] => Array ( [0] => 1 [1] => 5 [2] => 9 )
    [6] => Array ( [0] => 1 [1] => 6 [2] => 7 )
    [7] => Array ( [0] => 1 [1] => 6 [2] => 8 )
    [8] => Array ( [0] => 1 [1] => 6 [2] => 9 )
    [9] => Array ( [0] => 2 [1] => 4 [2] => 7 )
   [10] => Array ( [0] => 2 [1] => 4 [2] => 8 )
   [11] => Array ( [0] => 2 [1] => 4 [2] => 9 )
   [12] => Array ( [0] => 2 [1] => 5 [2] => 7 )
   [13] => Array ( [0] => 2 [1] => 5 [2] => 8 )
   [14] => Array ( [0] => 2 [1] => 5 [2] => 9 )
   [15] => Array ( [0] => 2 [1] => 6 [2] => 7 )
   [16] => Array ( [0] => 2 [1] => 6 [2] => 8 )
   [17] => Array ( [0] => 2 [1] => 6 [2] => 9 )
   [18] => Array ( [0] => 3 [1] => 4 [2] => 7 )
   [19] => Array ( [0] => 3 [1] => 4 [2] => 8 )
   [20] => Array ( [0] => 3 [1] => 4 [2] => 9 )
   [21] => Array ( [0] => 3 [1] => 5 [2] => 7 )
   [22] => Array ( [0] => 3 [1] => 5 [2] => 8 )
   [23] => Array ( [0] => 3 [1] => 5 [2] => 9 )
   [24] => Array ( [0] => 3 [1] => 6 [2] => 7 )
   [25] => Array ( [0] => 3 [1] => 6 [2] => 8 )
   [26] => Array ( [0] => 3 [1] => 6 [2] => 9 )
)

如果你想得到总和，你可以简单地使用 array_map在数组上，调用 array_sum获取每个元素的总和:

$sums = array_map(function ($v) { return array_sum($v);}, $combinations);

输出:

Array (
    [0] => 12
    [1] => 13
    [2] => 14
    [3] => 13
    [4] => 14
    [5] => 15
    [6] => 14
    [7] => 15
    [8] => 16
    [9] => 13
   [10] => 14
   [11] => 15
   [12] => 14
   [13] => 15
   [14] => 16
   [15] => 15
   [16] => 16
   [17] => 17
   [18] => 14
   [19] => 15
   [20] => 16
   [21] => 15
   [22] => 16
   [23] => 17
   [24] => 16
   [25] => 17
   [26] => 18 
)

Demo on 3v4l.org