ruby – 大型套装的独特排列
发布时间:2020-12-17 04:00:58 所属栏目:百科 来源:网络整理
导读:我正试图找到一种方法来为y长度的x值生成唯一的排列.我想要做的是: [0,1].unique_permutations(15)# = [[0,0],# [0,1],1,# ... massive snip# [1,# [1,# ... massive snip# [0, 要清楚,我知道这是可能的: [0,1].permutation.count# = 720[0,1].permutation
我正试图找到一种方法来为y长度的x值生成唯一的排列.我想要做的是:
[0,1].unique_permutations(15) # => [[0,0],# [0,1],1,# ... massive snip # [1,# [1,# ... massive snip # [0, 要清楚,我知道这是可能的: [0,1].permutation.count # => 720 [0,1].permutation.to_a.uniq.count # => 20 但这与我正在寻找的并不完全相同,对于长列表而言,性能变得不切实际: [1,1].permutation.count # => 479001600 (after a long wait...) [1,1].permutation.to_a.uniq.count # => 1 (didn't actually run this - answer is obvious) 我能找到的最近的东西是this answer for python,但遗憾的是我不知道python并且无法弄清楚如何将它移植到Ruby. 我确定这类问题还有其他算法,但我真的很想把它保存在Ruby中. 解决方法
你正在寻找的是一组自身的n-ary笛卡尔乘积(在你的例子中,n = 15超过set [0,1].)这与你后面引用的#permutation列表不同.题.
此列表的大小随n呈指数增长.除了微小的n之外,实际实现它是不切实际的.你可以改用发电机(原谅我生锈的ruby): class Array def cartesian_power(n) current = [0] * n last = [size - 1] * n loop do yield current.reverse.collect { |i| self[i] } break if current == last (0...n).each do |index| current[index] += 1 current[index] %= size break if current[index] > 0 end end end end 然后: >> [0,1].cartesian_power(3) { |l| puts l.inspect } [0,0] [0,1] [0,1] [1,0] [1,1] => nil >> %w{a b c}.cartesian_power(2) { |l| puts l.inspect } ["a","a"] ["a","b"] ["a","c"] ["b","a"] ["b","b"] ["b","c"] ["c","a"] ["c","b"] ["c","c"] => nil (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |