如何在C＃中生成没有重复（字符）的随机长度为26的字符串？

Question

 Random rnd = new Random();
        string[] coupon = new string[26];
        for (int i = 0; i < coupon.Length; i++)
        {
            coupon[i] = GenerateCoupon(26, rnd);
        }
        textBox1 .Text=(String.Join(Environment.NewLine, coupon));

***** Function***********\\

  public static string GenerateCoupon(int length, Random random)
    {
        string characters = "abcdefghijklmnopqrstuvwxyz";
        StringBuilder result = new StringBuilder(length);
        {
            result.Append(characters[random.Next(characters.Length)]);

            return result.ToString();
        }

Strings that code generates :

kheeuasampxqxmoohcrufznugp

vrlncvbftinynhdufjdikacjsi

vblltkxeeapymbprtgaiojqkte

qyfvpcvtazuiodbidcfgwcssgw

ijtlkbrpuyzilndsaqxlrxhggo

emhngmostlapotqziciursddcc

vvflcnewwehgsntstrskbduroe

But i need a code that generates the string that have 26 character length and no duplicate character :

abcdefghijklmnopqrstuvwxyz

zyxwvutsrqponmlkjihgfedcba

mnbvcxzasdfghjklpoiuytrewq

qazwsxedcrfvtgbyhnujmikolp

bhuijnmkoplqazxswedcvfrtgb

Answer 1

You can use Fisher-Yates shuffle , or simply

Random rnd = new Random();
var newstr = String.Concat("abcdefghijklmnopqrstuvwxyz".OrderBy(x => rnd.Next()));

Answer 2

查看一些随机播放算法，并将其应用于您的字符集。

Answer 3

You can just have a string of az . For example: "abcdefghijklmnopqrstuvwxyz".

After that, randomly shuffle the string based on the index of each letter.