获取所有（下一个）子集或排列（或其他东西）

Question

我不太确定这里使用的词，所以如果我使用了错误的术语，请原谅。

我正在尝试创建一个函数来获取给出当前字符串和允许字符的字符串的下一个排列。

例如

<pre>
<?php
$current = '';
$allowed = 'ab';

function next(&$current, &$allowed) {
    // This is where I need help
}

echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";
echo next($current, $allowed) . "\n";

，Should return

a
b
aa
ab
ba
bb
aaa
aab
aba
abb
baa
bab
bba
bbb
aaaa

...等等，

我正在尝试在 PHP 和 JavaScript 中执行此操作，因此我将不胜感激任何一种语言的帮助。

Answer 1

function nextPermutation(&$current, $allowed) {
    if (empty($current)) {
        $current = $allowed[0];
    } else {
        for ($i = strlen($current) - 1; $i >= 0; $i--) {
            $index = strpos($allowed, $current[$i]);
            if ($index < strlen($allowed) - 1) {
                $current[$i] = $allowed[$index + 1];
                break;
            } else {
                $current[$i] = $allowed[0];
                if ($i == 0) {
                    $current = $allowed[0] . $current;
                    break;
                }
            }
        }
    }
    return $current;
}