这是一个递归解决方案(使用函数式编程风格):
public static IEnumerable<IEnumerable<int>> GetCombinations(IEnumerable<int> limits)
{
if (limits.Any() == false)
{
// Base case.
yield return Enumerable.Empty<int>();
}
else
{
int first = limits.First();
IEnumerable<int> remaining = limits.Skip(1);
IEnumerable<IEnumerable<int>> tails = GetCombinations(remaining);
for (int i = 0; i < first; ++i)
foreach (IEnumerable<int> tail in tails)
yield return Yield(i).Concat(tail);
}
}
// Per http://stackoverflow.com/q/1577822
public static IEnumerable<T> Yield<T>(T item)
{
yield return item;
}
样品用途:
var sequences = GetCombinations(new [] { 5, 3, 2, 4 /* ... */ });
foreach (var sequence in sequences)
Console.WriteLine(string.Join(", ", sequence));
/* Output:
0, 0, 0, 0
0, 0, 0, 1
0, 0, 0, 2
0, 0, 0, 3
0, 0, 1, 0
0, 0, 1, 1
0, 0, 1, 2
0, 0, 1, 3
0, 1, 0, 0
0, 1, 0, 1
0, 1, 0, 2
... */
对于 OP 的特定场景(将数组添加到collection
):
var sequences = GetCombinations(new [] { 10, 20, 10 });
collection.AddRange(sequences.Select(s => s.ToArray()));