我想有效地生成的基于数字的开始列表上的数字的组合的唯一列表。

I would like to efficiently generate a unique list of combinations of numbers based on a starting list of numbers.

例如启动列表= [1,2,3,4,5]，但该算法应适用于[1,2,3 ... N]

example start list = [1,2,3,4,5] but the algorithm should work for [1,2,3...n]

结果=

[1]，[2]，[3]，[4]，[5] [1,2]，[1,3]，[1,4]，[1,5] [1,2,3]，[1,2,4]，[1,2,5] [1,3,4]，[1,3,5]，[1,4,5] [2,3]，[2,4]，[2,5] [2,3,4]，[2,3,5] [3,4]，[3,5] [3,4,5] [4,5]

[1],[2],[3],[4],[5] [1,2],[1,3],[1,4],[1,5] [1,2,3],[1,2,4],[1,2,5] [1,3,4],[1,3,5],[1,4,5] [2,3],[2,4],[2,5] [2,3,4],[2,3,5] [3,4],[3,5] [3,4,5] [4,5]

请注意。我不想重复的组合，虽然我可以和他们住在一起，例如，在上面的例子中，我并不真正需要的组合[1,3,2]，因为它已经present为[1,2,3]

Note. I don't want duplicate combinations, although I could live with them, eg in the above example I don't really need the combination [1,3,2] because it already present as [1,2,3]

推荐答案

没有为你问一个名称。这就是所谓的电源设置。

There is a name for what you're asking. It's called the power set.

谷歌搜索权力集法使我这个递归解决方案。

Googling for "power set algorithm" led me to this recursive solution.

有关的仇敌：

def powerset(set)
  return [set] if set.empty?

  p = set.pop
  subset = powerset(set)  
  subset | subset.map { |x| x | [p] }
end