需求背景：计算出数组['A','B','C','D']各种排列组合，希望得到的是数据如下图

直接上代码：
 

 private function finish_combination($array, &$groupResult = [], $splite = ','){$result = [];$finish_result = [];$this->diffArrayItems($array, $result);foreach ($result as $value) {$items = $this->array_combination($value, $splite);if (!isset($groupResult[count($value)])) $groupResult[count($value)] = [];$groupResult[count($value)] = array_merge($groupResult[count($value)], $items);$finish_result = array_merge($finish_result, $items);}return $finish_result;}
// 不同数组的值进行排列组合例如：['A','B','C'],排列如何出所有的可能性private function array_combination($arr, $splite = ','){$len = count($arr);if ($len == 1) {return $arr;}$result = array();for ($i = 0; $i < $len; $i++) {$tmp_arr = $arr;unset($tmp_arr[$i]);$tmp_arr = array_values($tmp_arr);$tmp_result = $this->array_combination($tmp_arr, $splite);foreach ($tmp_result as $val) {$val .= $splite . $arr[$i];$result[] = $val;}}return $result;}private function getCombinations($input, $length, $start = 0, $current = array(), &$result = array()){if (count($current) === $length) {$result[] = $current;return;}for ($i = $start; $i < count($input); $i++) {$current[] = $input[$i];$this->getCombinations($input, $length, $i + 1, $current, $result);array_pop($current);}}
// 获取不同个数的数组例如['A']，['B'],['C'],['D'],['A''B'],['A''c']......['A', 'B', 'C', 'D']private function diffArrayItems($input, &$result){for ($i = 1; $i <= count($input); $i++) {$this->getCombinations($input, $i, 0, array(), $result);}}//直接调用
$result = $this->finish_combination(['A', 'B', 'C', 'D']);
var_dump($result);

// 排列组合了所有的数据后，一般情况下，我们都想着验证下个数是否正确，
例如：
取 1 个元素的排列组合数为 P(4, 1) = 4! / (4 - 1)! = 4! / 3! = 4
取 2 个元素的排列组合数为 P(4, 2) = 4! / (4 - 2)! = 4! / 2! = 12
取 3 个元素的排列组合数为 P(4, 3) = 4! / (4 - 3)! = 4! / 1! = 24
取 4 个元素的排列组合数为 P(4, 4) = 4! / (4 - 4)! = 4! / 0! = 24
下面就需要另外一个函数

//此方法需要开启GMP扩展
private function nP($n){$sum = 0;for ($i = 1; $i <= $n; $i++) {$sum += gmp_fact($n) / gmp_fact($n - $i);}return intval($sum);}
// 然后直接调用
echo $this->nP(count('A', 'B', 'C', 'D']));