List <List <int >>的组合

我有一个这种类型的列表List>包含这个

List A = new List {1, 2, 3, 4, 5};
List B = new List {0, 1};
List C = new List {6};
List X = new List {....,....};

我希望拥有这样的所有组合

1-0-6
1-1-6
2-0-6
2-1-6
3-0-6

等等.

据你说这是使用Linq解决这个问题吗？

这与我给另一个问题的答案非常相似:

var combinations = from a in A
                   from b in B
                   from c in C
                   orderby a, b, c
                   select new List { a, b, c };

var x = combinations.ToList();

对于可变数量的输入,现在添加了泛型:

var x = AllCombinationsOf(A, B, C);

public static List> AllCombinationsOf(params List[] sets)
{
    // need array bounds checking etc for production
    var combinations = new List>();

    // prime the data
    foreach (var value in sets[0])
        combinations.Add(new List { value });

    foreach (var set in sets.Skip(1))
        combinations = AddExtraSet(combinations, set);

    return combinations;
}

private static List> AddExtraSet
     (List> combinations, List set)
{
    var newCombinations = from value in set
                          from combination in combinations
                          select new List(combination) { value };

    return newCombinations.ToList();
}

如果维度的数量是固定的,这只是SelectMany:

var qry = from a in A
          from b in B
          from c in C
          select new {A=a,B=b,C=c};

但是,如果维度的数量由数据控制,则需要使用递归:

static void Main() {
    List> outerList = new List>
    {   new List(){1, 2, 3, 4, 5},
        new List(){0, 1},
        new List(){6,3},
        new List(){1,3,5}
    };
    int[] result = new int[outerList.Count];
    Recurse(result, 0, outerList);
}
static void Recurse(int[] selected, int index,
    IEnumerable remaining) where TList : IEnumerable {
    IEnumerable nextList = remaining.FirstOrDefault();
    if (nextList == null) {
        StringBuilder sb = new StringBuilder();
        foreach (int i in selected) {
            sb.Append(i).Append(',');
        }
        if (sb.Length > 0) sb.Length--;
        Console.WriteLine(sb);
    } else {
        foreach (int i in nextList) {
            selected[index] = i;
            Recurse(selected, index + 1, remaining.Skip(1));
        }
    }
}

如何使用.Join方法生成组合的方法如何？

static void Main()
{
    List> collectionOfSeries = new List>
                                {   new List(){1, 2, 3, 4, 5},
                                    new List(){0, 1},
                                    new List(){6,3},
                                    new List(){1,3,5}
                                };
    int[] result = new int[collectionOfSeries.Count];

    List> combinations = GenerateCombinations(collectionOfSeries);

    Display(combinations); 
}

此方法GenerateCombinations(..)执行生成组合的主要工作.此方法是通用的,因此可用于生成任何类型的组合.

private static List> GenerateCombinations(
                                List> collectionOfSeries)
{
    List> generatedCombinations = 
        collectionOfSeries.Take(1)
                          .FirstOrDefault()
                          .Select(i => (new T[]{i}).ToList())                          
                          .ToList();

    foreach (List series in collectionOfSeries.Skip(1))
    {
        generatedCombinations = 
            generatedCombinations
                  .Join(series as List,
                        combination => true,
                        i => true,
                        (combination, i) =>
                            {
                                List nextLevelCombination = 
                                    new List(combination);
                                nextLevelCombination.Add(i);
                                return nextLevelCombination;
                            }).ToList();

    }

    return generatedCombinations;
}

显示助手..

private static void Display(List> generatedCombinations)
{
    int index = 0;
    foreach (var generatedCombination in generatedCombinations)
    {
        Console.Write("{0}\t:", ++index);
        foreach (var i in generatedCombination)
        {
            Console.Write("{0,3}", i);
        }
        Console.WriteLine();
    }
    Console.ReadKey();
}