问题 {#heading}

我有一组字符："ab"，我需要每个给定长度的组合和排列。

如果我给定长度为3，我需要像这样的集合：
a，
b，
aa，
ab，
ba，
bb，
aaa，
aab，
aba，
abb，
baa，
bab，
bba，
bbb

以此类推。

我找到了一个相对不错的解决方案（https://stackoverflow.com/a/3640224/18535919），但我认为它有点复杂。过去我找到过一个好的解决方案，但现在找不到了。

static int maxlength = 3;
static string ValidChars = "abc";
static void Dive(string prefix, int level)
{
       level += 1;
       foreach (char c in ValidChars)
       {
            Console.WriteLine(prefix + c);
            if (level < maxlength)
            {
                Dive(prefix + c, level);
            }
      }
}

英文:

I have a set of characters: "ab", I need every combinations and permutations of it to given length.
What I need sets like these if I give the length of 3:
a,
b,
aa,
ab,
ba,
bb,
aaa,
aab,
aba,
abb,
baa,
bab,
bba,
bbb

and so on.

I found a sort of good solution (https://stackoverflow.com/a/3640224/18535919), but I think it is a bit complicated. I have found a good solution in the past but I can not find it again.

static int maxlength = 3;
static string ValidChars = &quot;abc&quot;;
static void Dive(string prefix, int level)
{
       level += 1;
       foreach (char c in ValidChars)
       {
            Console.WriteLine(prefix + c);
            if (level &lt; maxlength)
            {
                Dive(prefix + c, level);
            }
      }
}

答案1 {#1}

得分: 1

以下是翻译好的部分：

你当前的代码看起来足够简单，但如果你正在寻找一个不使用递归的通用方法，你可以尝试像这样的东西（坦白地说，这更加复杂，但可以在一般情况下使用）：

示例

using System.Linq;

...

private static IEnumerable<T[]> SolveMe<T>(int maxLength, IEnumerable<T> chars) {
  if (maxLength < 0)
    throw new ArgumentOutOfRangeException(nameof(maxLength));

  if (chars is null)
    throw new ArgumentNullException(nameof(chars));

  var letters = chars.Distinct().ToArray();

  if (letters.Length == 0)
     yield break;

  for (var agenda = new Queue<T[]>(new[] { Array.Empty<T>() }); 
           agenda.Peek().Length < maxLength;) {
    var current = agenda.Dequeue();

    foreach (var letter in letters) {
      var next = current.Append(letter).ToArray();

      agenda.Enqueue(next);

      yield return next;
    }
  }
}

演示：

// 所有解决方案作为列表
List<char[]> result = SolveMe(3, "ab").ToList();

var report = string.Join(", ", result.Select(item => string.Concat(item)));

Console.Write(report);

输出：

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

英文:

Your current code looks simple enough, however if you are looking for generalized approach which doesn't use recursion you can try something like this (frankly speaking it's more complicated , but can be used in general case):

Fiddle

using System.Linq;

...


private static IEnumerable\&lt;T\[\]\&gt; SolveMe\&lt;T\&gt;(int maxLength, IEnumerable\&lt;T\&gt; chars) {
if (maxLength \&lt; 0)
throw new ArgumentOutOfRangeException(nameof(maxLength));


if (chars is null)
throw new ArgumentNullException(nameof(chars));


var letters = chars.Distinct().ToArray();


if (letters.Length == 0)
yield break;


for (var agenda = new Queue\&lt;T\[\]\&gt;(new\[\] { Array.Empty\&lt;T\&gt;() });
agenda.Peek().Length \&lt; maxLength;) {
var current = agenda.Dequeue();


    foreach (var letter in letters) {
      var next = current.Append(letter).ToArray();

      agenda.Enqueue(next);

      yield return next;
    }



`}
}
`

Demo:

// All solutions as a list
List&lt;char[]&gt; result = SolveMe(3, &quot;ab&quot;).ToList();

var report = string.Join(\&quot;, \&quot;, result.Select(item =\&gt; string.Concat(item)));

`Console.Write(report);
`

Output:

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

答案2 {#2}

得分: 1

你可以使用递归技巧来生成给定字符的所有组合和排列。请查看下面的代码：

class Program
{
    static void Main(string[] args)
    {
        string characters = "ab";
        int maxLength = 3;
        var output = GenerateCombinations(characters, maxLength, "");
        Console.WriteLine(string.Join(", ", output.Select(item => string.Concat(item))));
    }

    static IEnumerable<string> GenerateCombinations(string characters, int maxLength, string current)
    {
        if (current.Length > 0)
        {
            yield return current;
        }

        if (current.Length < maxLength)
        {
            foreach (char c in characters)
            {
                foreach (string result in GenerateCombinations(characters, maxLength, current + c))
                {
                    yield return result;
                }
            }
        }
    }
}

输出结果: a, aa, aaa, aab, ab, aba, abb, b, ba, baa, bab, bb, bba, bbb 英文:

You can achieve this using recursive techniques to generate all the combinations and permutations of the given characters. Check the below code:

class Program
{
    static void Main(string[] args)
    {
        string characters = &quot;ab&quot;;
        int maxLength = 3;
        var output = GenerateCombinations(characters, maxLength, &quot;&quot;);
        Console.WriteLine(string.Join(&quot;, &quot;, output.Select(item =&gt; string.Concat(item))));
    }

    static IEnumerable&amp;lt;string&amp;gt; GenerateCombinations(string characters, int maxLength, string current)
    {
        if (current.Length &amp;gt; 0)
        {
            yield return current;
        }

        if (current.Length &amp;lt; maxLength)
        {
            foreach (char c in characters)
            {
                foreach (string result in GenerateCombinations(characters, maxLength, current + c))
                {
                    yield return result;
                }
            }
        }
    }



`}
`

Output: a, aa, aaa, aab, ab, aba, abb, b, ba, baa, bab, bb, bba, bbb