列出不同字符组合的排列

Listing permutations of different combination of characters

我找到的最接近的 SO 主题在这里:Listing all permutations of a string/integer

但是我如何将它用于字符串中每个位置的不同字符集?

举个例子:我指定字符串长度为“3”。前两个位置应为 "a" 或 "b",但最后一个位置应为“1”或“2”,例如:

aa1
ba1
ab1
bb1
aa2
ab2
ba2
bb2

如果长度是固定的,您可以使用这个简单的查询来创建 cartesian product:

string chars = "ab";
int[] digits = { 1, 2 };
var query = from c1 in chars 
            from c2 in chars 
            from d1 in digits 
            select string.Format("{0}{1}{2}", c1, c2, d1);
string[] possibleCombinations = query.ToArray();

结果:

aa1
aa2
ab1
ab2
ba1
ba2
bb1
bb2

编辑:对于它的价值,按要求使用 lambda(查询语法更具可读性):

possibleCombinations = chars
    .SelectMany(c1 => chars
        .SelectMany(c2 => digits
            .Select(d1 => string.Format("{0}{1}{2}", c1, c2, d1))))
    .ToArray();

如果您需要一种处理动态长度的方法,您可以看看这个:

Dynamic Generation of All Possible Combinations of Index of an Array

使用此代码:

public static List<string> GenerateCombinations(char[][] characters)
{
    var combinations = new List<string>();
    GenerateCombinations(0, characters, new char[characters.GetLength(0)], combinations);
    return combinations;
}

private static void GenerateCombinations(int level, char[][] characters, char[] current, List<string> combinations)
{
    if (level == characters.GetLength(0))
    {
        combinations.Add(new string(current));
        return;
    }

    foreach (var character in characters[level])
    {
        current[level] = character;
        GenerateCombinations(level + 1, characters, current, combinations);
    }
}

使用示例:

public static void Main()
{
    var characters = new[]
                     {
                         new[] { 'a', 'b' },
                         new[] { 'a', 'b' },
                         new[] { '1', '2' }
                     };

    var combinations = GenerateCombinations(characters);
    foreach (var combination in combinations)
    {
        Console.WriteLine(combination);
    }
}

输出:

aa1
aa2
ab1
ab2
ba1
ba2
bb1
bb2