场景: 只能用6位数, 要表示0~百亿区间的数. etc. xabdex => 2000000000.

理解进制

二进制, 逢2进1. 101 => 1x2^2 + 0x2^1 + 1x2^0 = 5 二进制3为才能表示10进制的1位数.
16进制, 逢16进1. 101 => 1x16^2 + 0x16^1 + 1x16^0 = 257 16进制101表示到了257.

64进制, 逢64进1. 101 => 1x64^2 + 0x64^1 + 1x64^0 = 4160 64进制三位表示到了4位数, 就是我们想要的.

实现(php)

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class UIDUtils
{

    /**
     * 用户唯一表示长度
     */
    const CODE_LENGTH = 6;


    /**
     * if number is negative prefix with zero
     *
     */
    private static $negative = 0;

    /**
     * this character presentation zero
     */
    private static $zeroCharacter = [ 'J', 'q', 's', 'T', 'U', 'E'];

    /**
     * code dictionary
     */
    private static $dictionary = [
          0 => 'K',
          1 => 'v',
          2 => 'F',
          3 => 'm',
          4 => 'u',
          5 => 'B',
          6 => 'S',
          7 => 'n',
          8 => 'Y',
          9 => 'e',
          10 => 'O',
          11 => 'd',
          12 => 'A',
          13 => 'j',
          14 => 'x',
          15 => 'y',
          16 => 'V',
          17 => 'X',
          18 => 'b',
          19 => 'W',
          20 => 'Q',
          21 => 'P',
          22 => 'z',
          23 => 't',
          24 => 'r',
          25 => 'Z',
          26 => 'R',
          27 => 'w',
          28 => 'c',
          29 => 'p',
          30 => 'I',
          31 => 'G',
          32 => 'k',
          33 => 'C',
          34 => 'h',
          35 => 'f',
          36 => 'l',
          37 => 'i',
          38 => 'o',
          39 => 'N',
          40 => 'L',
          41 => 'M',
          42 => 'D',
          43 => 'a',
          44 => 'g',
          45 => 'H',
    ];

    /**
     * encrypt function get code
     *
     * @param $number
     * @return string
     * @throws \Error
     */
    public static function encode($number)
    {
        $encodeString = '';
        $numberType = $number < 0  ? true : false;

        $dictionaryLength = count(self::$dictionary);

        while($number > $dictionaryLength)
        {
            $encodeString = self::$dictionary[abs($number % $dictionaryLength)] . $encodeString;
            $number = floor($number / $dictionaryLength);
        }
        if($number > 0)
            $encodeString = self::$dictionary[$number] . $encodeString;

        $encodeStringLength = strlen($encodeString);
        $padStringLength = self::CODE_LENGTH - $encodeStringLength;


        // if number is negative number and encode length is six. number overflow
        if(($numberType && $padStringLength < 1) || (!$numberType && $padStringLength < 0))
            throw new \Error("加密数字溢出....");

        // if number is positive and encode length is six. just return it.
        if($padStringLength == 0)
            return $encodeString;

        // if number is positive and encode length small than six. pad random character left.
        shuffle(self::$zeroCharacter);

        if($numberType)
            $resultEncodeString = self::$negative . implode(array_slice(self::$zeroCharacter, 0, $padStringLength + 1)) . $encodeString;
        else
            $resultEncodeString = implode(array_slice(self::$zeroCharacter, 0, $padStringLength)) . $encodeString;

        return $resultEncodeString;
    }

    /**
     *  decode string to number value
     *
     * @param $string
     * @return int
     * @throws \Error
     */
    public static function decode($string)
    {
        $numberValue = 0;
        $dictionaryLength = count(self::$dictionary);
        $decodeArr = array_reverse(str_split($string));

        if(count($decodeArr) != self::CODE_LENGTH)
            throw new \Error("字符长度错误.");

        $dictionaryNumberMap = array_flip(self::$dictionary);
        foreach($decodeArr as $key => $value)
        {
            if(in_array($value, self::$zeroCharacter))
                continue;
            $numberValue += $dictionaryNumberMap[$value] * pow($dictionaryLength, $key);
        }
        return $numberValue;
    }

$number = 10000000;

$string = UIDUtils::encode($number);
var_dump($string);

$number = UIDUtils::decode($string);
var_dump($number);

思路很重要, 实现很简单.