c# - 迭代非方(某种)矩阵

Question

我有一个这样的字典(样本数据因此没有意义):

Dictionary<char, string[]> codes = new Dictionary<char, string[]>();

string[] 是字典键的可能替换数组。

正在用一些示例数据填充字典...

codes.Add("A", new string[] {"噅噅", "裧", "頖", "11"});
codes.Add("B", new string[] {"垥", "2", "鉠"});
codes.Add("C", new string[] {"33", "韎"});
codes.Add("D", new string[] {"櫋", "緟", "嘕", "鈖", "灡", "犅"});
...
codes.Add("T", new string[] {"濇", "汫", "岕", "5"});
...

现在让我们“编码”以下单词:

    char[] charArray = "act".ToCharArray();
    foreach (char c in charArray) {
       string[] replacements = codes[c].Where(x => !String.IsNullOrEmpty(x)).ToArray();//removing empty elements

       ...
    }

我现在不知道下一步该做什么，我想要一个所有可能组合的列表，它应该返回一个像这样的列表(对于单词“act”):

噅噅韎5

裧33濇

裧33汫

裧33岕

裧335

裧韎濇

裧韎汫

裧韎岕

...

由于 stackoverflow 的垃圾邮件过滤器，无法显示所有组合...

Answer 1

帖子标题具有误导性。如果我理解正确，您想使用 codes 作为输入字符替换来生成所有组合。

我已经回答了一个类似的问题 ( String combinations with multi-character replacement )，但是由于 string[] 部分，我无法直接重用 Looking at each combination in jagged array 中的方便代码，因此我将针对您的情况使用相同的算法:

public static class Algorithms
{
    public static IEnumerable<string> GetCombinations(this string input, Dictionary<char, string[]> replacements)
    {
        var map = new string[input.Length][];
        for (int i = 0; i < map.Length; i++)
        {
            var c = input[i];
            string[] r;
            map[i] = replacements.TryGetValue(c, out r)
                && (r = r.Where(s => !string.IsNullOrEmpty(s)).ToArray()).Length > 0 ?
                r : new string[] { c.ToString() };
        }
        int maxLength = map.Sum(output => output.Max(s => s.Length));
        var buffer = new char[maxLength];
        int length = 0;
        var indices = new int[input.Length];
        for (int pos = 0, index = 0; ;)
        {
            for (; pos < input.Length; pos++, index = 0)
            {
                indices[pos] = index;
                foreach (var c in map[pos][index])
                    buffer[length++] = c;
            }
            yield return new string(buffer, 0, length);
            do
            {
                if (pos == 0) yield break;
                index = indices[--pos];
                length -= map[pos][index].Length;
            }
            while (++index >= map[pos].Length);
        }
    }
}

测试:

var codes = new Dictionary<char, string[]>();
codes.Add('A', new string[] { "噅噅", "裧", "頖", "11" });
codes.Add('B', new string[] { "垥", "2", "鉠" });
codes.Add('C', new string[] { "33", "韎" });
codes.Add('D', new string[] { "櫋", "緟", "嘕", "鈖", "灡", "犅" });
//...
codes.Add('T', new string[] { "濇", "汫", "岕", "5" });
//...

var test = "ACT".GetCombinations(codes).ToList();