【发布时间】：2012-04-21 04:54:44
【问题描述】：
我需要找到 N 组 X 长度且不重复且按特定顺序的所有可能组合，例如
input: [["A"], ["B"], ["C"]]
output: [["A","B","C"],["A","B"],["A","C"],["B","C"],["A"],["B"],["C"]]
规则：
- 分区的数量或大小不固定。
- 每个组合中的每个分区只有一个成员。
- 成员越多的组合优先级越高。
- 输入中较早的成员的优先级高于后来的成员。
另一个更大集合的例子：
input: [["A","B"],["C","D","E"],["F"]]
output: [["A","C","F"],["A","D","F"],["A","E","F"],["B","C","F"],["B","D","F"],["B","E","F"],["A","C"],["A","D"],["A","E"],["B","C"],["B","D"],["B","E"],["A","F"],["B","F"],["C","F"],["D","F"],["E","F"],["A"],["B"],["C"],["D"],["E"],["F"]]
通过将幂集函数的输出与笛卡尔积函数相结合，我设法获得了我想要的输出，但生成的代码不是很简洁或漂亮。我想知道这是否可以通过递归更好地完成？
这是我已经拥有的：
$test = json_decode('[["A"]]');
$test1 = json_decode('[["A"], ["B"], ["C"]]');
$test2 = json_decode('[["A", "B"], ["C", "D", "E"], ["F"]]');
/**
* Returns a power set of the input array.
*/
function power_set($in, $minLength = 1) {
$count = count($in);
$members = pow(2,$count);
$return = array();
for ($i = 0; $i < $members; $i++) {
$b = sprintf("%0".$count."b",$i);
$out = array();
for ($j = 0; $j < $count; $j++) {
if ($b[$j] == '1') {
$out[] = $in[$j];
}
}
if (count($out) >= $minLength) {
$return[] = $out;
}
}
return $return;
}
/**
* Returns the cartesian product of the input arrays.
*/
function array_cartesian() {
$_ = func_get_args();
if(count($_) == 0) {
return array(array());
}
$a = array_shift($_);
$c = call_user_func_array(__FUNCTION__, $_);
$r = array();
foreach($a as $v) {
foreach($c as $p) {
$r[] = array_merge(array($v), $p);
}
}
return $r;
}
/**
* Used with usort() to sort arrays by length, desc.
* If two arrays are the same length, then a sum of
* their keys is taken, with lower values coming first.
*/
function arraySizeDesc($a, $b) {
if(count($a) === count($b)) {
if(array_sum($a) === array_sum($b)) {
return 0;
}
return (array_sum($a) > array_sum($b)) ? 1 : -1;
}
return (count($a) < count($b)) ? 1 : -1;
}
/**
* Calculates a powerset of the input array and then uses
* this to generate cartesian products of each combination
* until all possible combinations are aquired.
*/
function combinations($in) {
$out = array();
$powerSet = power_set(array_keys($in));
usort($powerSet, 'arraySizeDesc');
foreach($powerSet as $combination) {
if(count($combination) < 2) {
foreach($in[$combination[0]] as $value) {
$out[] = array($value);
}
} else {
$sets = array();
foreach($combination as $setId) {
$sets[] = $in[$setId];
}
$out = array_merge($out, call_user_func_array('array_cartesian', $sets));
}
}
return $out;
}
echo "input: ".json_encode($test2);
echo "<br />output: ".json_encode(combinations($test2));
我意识到输出的大小可以非常迅速地增长，但输入通常应该只包含 1-5 个 1-50 个成员的集合，因此不需要处理大量集合。
【问题讨论】：
标签： php algorithm recursion set cartesian