获取所有(下一个)子集或排列(或其他)

时间:2011-02-15 12:14:04

标签: php javascript set permutation

我不太确定在这里使用这个词,如果我使用错误的术语,请原谅。

我正在尝试创建一个函数来获取字符串的下一个排列,给出当前字符串和一串允许的字符。

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";

应该返回

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,因此我将非常感谢您使用这两种语言提供帮助。

1 个答案:

答案 0 :(得分:0)

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;
}