在Linq(C#)中将多个不同大小的集合动态交叉连接在一起

您可以创建如下的扩展方法：

public static class EnumerableExtensions
{
    public static IEnumerable<TValue []> Permutations<TKey, TValue>(this IEnumerable<TKey> keys, Func<TKey, IEnumerable<TValue>> selector)
    {
        var keyArray = keys.ToArray();
        if (keyArray.Length < 1)
            yield break;
        TValue [] values = new TValue[keyArray.Length];
        foreach (var array in Permutations(keyArray, 0, selector, values))
            yield return array;
    }

    static IEnumerable<TValue []> Permutations<TKey, TValue>(TKey [] keys, int index, Func<TKey, IEnumerable<TValue>> selector, TValue [] values)
    {
        Debug.Assert(keys.Length == values.Length);

        var key = keys[index];
        foreach (var value in selector(key))
        {
            values[index] = value;
            if (index < keys.Length - 1)
            {
                foreach (var array in Permutations(keys, index+1, selector, values))
                    yield return array;
            }
            else
            {
                yield return values.ToArray(); // Clone the array;
            }
        }
    }
}

例如,它可以像：

public static void TestPermutations()
    {
        int [][] seqence = new int [][]
        {
            new int [] {1, 2, 3},
            new int [] {101},
            new int [] {201},
            new int [] {301, 302, 303},
        };

        foreach (var array in seqence.Permutations(a => a))
        {
            Debug.WriteLine(array.Aggregate(new StringBuilder(), (sb, i) => { if (sb.Length > 0) sb.Append(","); sb.Append(i); return sb; }));
        }
    }

并产生以下输出：

1,101,201,301
1,101,201,302
1,101,201,303
2,101,201,301
2,101,201,302
2,101,201,303
3,101,201,301
3,101,201,302
3,101,201,303

那是你要的吗？