如何将 int64 快速转换（重新编码）为字符限制字符串？

Question

我正在寻找一种方法来将仅由数字 [0,1,2,...8,9] 组成的字符串重新编码为另一个具有不同字符集的字符串，例如数字和字母 [0,1 ,...9,a,b,...z]（作为示例）。结果字符串通常应该比纯数字字符串短。该方法应该是可逆的并且应该是快速的。 C# 答案中的源代码将不胜感激，但也欢迎提出一般性想法；-) 示例：

Input:  "1234567890123"
Output: "ar4cju7d"

应用于"ar4cju7d"的反转方法的输出应该是“1234567890123.

Answer 1

如果您的输入始终是 Int64，那么您可以检查 Base36 encoding. And here's a sample implementation:

// Edit: Slightly updated on 2011-03-29

/// <summary>
/// A Base36 De- and Encoder
/// </summary>
public static class Base36
{
    private const string CharList = "0123456789abcdefghijklmnopqrstuvwxyz";

    /// <summary>
    /// Encode the given number into a Base36 string
    /// </summary>
    /// <param name="input"></param>
    /// <returns></returns>
    public static String Encode(long input)
    {
        if (input < 0) throw new ArgumentOutOfRangeException("input", input, "input cannot be negative");

        char[] clistarr = CharList.ToCharArray();
        var result = new Stack<char>();

        while (input != 0)
        {
            result.Push(clistarr[input % 36]);
            input /= 36;
        }

        return new string(result.ToArray());
    }

    /// <summary>
    /// Decode the Base36 Encoded string into a number
    /// </summary>
    /// <param name="input"></param>
    /// <returns></returns>
    public static Int64 Decode(string input)
    {
        var reversed = input.ToLower().Reverse();
        long result = 0;
        int pos = 0;

        foreach (char c in reversed)
        {
            result += CharList.IndexOf(c) * (long)Math.Pow(36, pos);
            pos++;
        }

        return result;
    }
}

可以应用于您的示例：

long input = 1234567890123;
string encoded = Base36.Encode(input); // yields "fr5hugnf"
long originalInput = Base36.Decode(encoded); // yields 1234567890123

如何将 int64 快速转换（重新编码）为字符限制字符串？

How to fast convert (recode) int64 to character-limited String?

c#

string

character-encoding

recode