一尘不染

获取PHP数组的所有排列?

php

给定一个PHP字符串数组,例如:

['peter', 'paul', 'mary']

如何生成此数组元素的所有可能排列?即:

peter-paul-mary
peter-mary-paul
paul-peter-mary
paul-mary-peter
mary-peter-paul
mary-paul-peter

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2020-05-26

共1个答案

一尘不染

function pc_permute($items, $perms = array()) {
if (empty($items)) {
echo join(‘ ‘, $perms) . “
“;
} else {
for ($i = count($items) - 1; $i >= 0; –$i) {
$newitems = $items;
$newperms = $perms;
list($foo) = array_splice($newitems, $i, 1);
array_unshift($newperms, $foo);
pc_permute($newitems, $newperms);
}
}
}

$arr = array('peter', 'paul', 'mary');

pc_permute($arr);

要么

function pc_next_permutation($p, $size) {
    // slide down the array looking for where we're smaller than the next guy
    for ($i = $size - 1; $p[$i] >= $p[$i+1]; --$i) { }

    // if this doesn't occur, we've finished our permutations
    // the array is reversed: (1, 2, 3, 4) => (4, 3, 2, 1)
    if ($i == -1) { return false; }

    // slide down the array looking for a bigger number than what we found before
    for ($j = $size; $p[$j] <= $p[$i]; --$j) { }

    // swap them
    $tmp = $p[$i]; $p[$i] = $p[$j]; $p[$j] = $tmp;

    // now reverse the elements in between by swapping the ends
    for (++$i, $j = $size; $i < $j; ++$i, --$j) {
         $tmp = $p[$i]; $p[$i] = $p[$j]; $p[$j] = $tmp;
    }

    return $p;
}

$set = split(' ', 'she sells seashells'); // like array('she', 'sells', 'seashells')
$size = count($set) - 1;
$perm = range(0, $size);
$j = 0;

do { 
     foreach ($perm as $i) { $perms[$j][] = $set[$i]; }
} while ($perm = pc_next_permutation($perm, $size) and ++$j);

foreach ($perms as $p) {
    print join(' ', $p) . "\n";
}
2020-05-26