从多个(n)列表生成所有组合

Question

编辑:我完全重做了我的问题,因为我已经找到了最简单的问题.感谢到目前为止的评论者让我思考根本问题.

public List<string> GetAllPossibleCombos(List<List<string>> strings)
{
    List<string> PossibleCombos = new List<string>();

    //????
    {
        string combo = string.Empty;
        // ????
        {
            combo += ????
        }
        PossibleCombos.Add(combo);
    }

    return PossibleCombos;
}

我需要弄清楚如何递归遍历每个List<string>并将每个列表中的1个字符串组合成一个组合string.不要过于担心格式化字符串,因为"实时"代码使用自定义对象.此外,请随意假设每个列表至少包含1个字符串,并且没有空值.

Answer 1

这是一个简单的非递归解决方案,它只是连接每个组合的元素:

public static List<string> GetAllPossibleCombos(List<List<string>> strings)
{
    IEnumerable<string> combos = new [] { "" };

    foreach (var inner in strings)
        combos = from c in combos
                 from i in inner
                 select c + i;

    return combos.ToList();
}

static void Main(string[] args)
{
    var x = GetAllPossibleCombos(
        new List<List<string>>{
            new List<string> { "a", "b", "c" },
            new List<string> { "x", "y" },
            new List<string> { "1", "2", "3", "4" }});
}

您可以将此概括为返回一个IEnumerable<IEnumerable<string>>,它允许调用者应用他们喜欢的任何操作将每个组合转换为字符串(string.Join如下所示).使用延迟执行枚举组合.

public static IEnumerable<IEnumerable<string>> GetAllPossibleCombos(
    IEnumerable<IEnumerable<string>> strings)
{
    IEnumerable<IEnumerable<string>> combos = new string[][] { new string[0] };

    foreach (var inner in strings)
        combos = from c in combos
                 from i in inner
                 select c.Append(i);

    return combos;
}

public static IEnumerable<TSource> Append<TSource>(
    this IEnumerable<TSource> source, TSource item)
{
    foreach (TSource element in source)
        yield return element;

    yield return item;
}

static void Main(string[] args)
{
    var combos = GetAllPossibleCombos(
        new List<List<string>>{
            new List<string> { "a", "b", "c" },
            new List<string> { "x", "y" },
            new List<string> { "1", "2", "3", "4" }});

    var result = combos.Select(c => string.Join(",", c)).ToList();
}

Answer 2

希望这可以帮助。

class NListBuilder

{
    Dictionary<int, List<string>> tags = new Dictionary<int, List<string>>();

    public NListBuilder()
    {
        tags.Add(1, new List<string>() { "A", "B", "C" });
        tags.Add(2, new List<string>() { "+", "-", "*" });
        tags.Add(3, new List<string>() { "1", "2", "3" });
    }

    public List<string> AllCombos
    {
        get
        {
            return GetCombos(tags);
        }
    }

    List<string> GetCombos(IEnumerable<KeyValuePair<int, List<string>>> remainingTags)
    {
        if (remainingTags.Count() == 1)
        {
            return remainingTags.First().Value;
        }
        else
        {
            var current = remainingTags.First();
            List<string> outputs = new List<string>();
            List<string> combos = GetCombos(remainingTags.Where(tag => tag.Key != current.Key));

            foreach (var tagPart in current.Value)
            {
                foreach (var combo in combos)
                {
                    outputs.Add(tagPart + combo);
                }
            }

            return outputs;
        }


    }
}

Answer 3

如果它对任何人有帮助，这里是道格拉斯GetAllPossibleCombos方法的方法语法版本。

 public static List<string> GetAllPossibleCombos(List<List<string>> strings)
 {
     IEnumerable<string> combos = new[] { "" };

     foreach (var inner in strings)
     {
         combos = combos.SelectMany(r => inner.Select(x => r + x));
     }

     return combos.ToList();
 }