TA貢獻1966條經(jīng)驗 獲得超4個贊
你想要的功率集$attributes？這就是你的問題所暗示的。
這里可以找到一個例子（引用完整性）
<?php
/**
* Returns the power set of a one dimensional array, a 2-D array.
* [a,b,c] -> [ [a], [b], [c], [a, b], [a, c], [b, c], [a, b, c] ]
*/
function powerSet($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 $return;
}
TA貢獻1853條經(jīng)驗 獲得超6個贊
使用php array_merge我們可以有一個很好的短powerSet函數(shù)
function powerSet($array) {
// add the empty set
$results = array(array());
foreach ($array as $element) {
foreach ($results as $combination) {
$results[] = array_merge(array($element), $combination);
}
}
return $results;
}
TA貢獻1795條經(jīng)驗 獲得超7個贊
這是一個回溯解決方案。
給定一個返回輸入集的所有L-lenght子集的函數(shù)，找到從L = 2到數(shù)據(jù)集輸入長度的所有L-lenght子集
<?php
function subsets($S,$L) {
$a = $b = 0;
$subset = [];
$result = [];
while ($a < count($S)) {
$current = $S[$a++];
$subset[] = $current;
if (count($subset) == $L) {
$result[] = json_encode($subset);
array_pop($subset);
}
if ($a == count($S)) {
$a = ++$b;
$subset = [];
}
}
return $result;
}
$S = [ 'A', 'B', 'C', 'D'];
$L = 2;
// L = 1 -> no need to do anything
print_r($S);
for ($i = 2; $i <= count($S); $i++)
print_r(subsets($S,$i));
舉報