列表的排列 - Haskell

Question

我想制作具有2个列表的子组的所有可能组合。这是一个功能：

getCombinations :: [a] -> [[a]]
getCombinations na = do
    a <- na
    b <- na
    [[a,b]]

如果将“abc”传递给此函数，则返回：

["aa","ab","ac","ba","bb","bc","ca","cb","cc"]

对同一方法的简单修改可以返回3个列表的组合，而不是两个。

getCombinations :: [a] -> [[a]]
getCombinations na = do
    a <- na
    b <- na
    c <- na
    [[a,b,c]]

将“abc”作为参数传递的结果：

["aaa","aab","aac","aba","abb","abc","aca","acb","acc",
"baa","bab","bac","bba","bbb","bbc","bca","bcb","bcc",
"caa","cab","cac","cba","cbb","cbc","cca","ccb","ccc"]

使其扩展到任意数量的列表的最简单方法是什么？以下是类型声明的样子：

getCombinations :: Int -> [a] -> [[a]]

Answer 1

你想要的是replicateM：

replicateM :: Int -> m a -> m [a]

定义很简单：

replicateM n = sequence . replicate n

所以在列表monad上sequence正在这里做真正的工作。

Answer 2

对于combination函数来到这里的人来说， k - 集合 S 的组合是<的一个子集em> k S 的不同元素，请注意订单并不重要。

从k元素中选择n元素等于从k - 1元素中选择n - 1元素，然后从k元素中选择n - 1元素。

使用这个递归定义，我们可以写：

combinations :: Int -> [a] -> [[a]]
combinations k xs = combinations' (length xs) k xs
  where combinations' n k' l@(y:ys)
          | k' == 0   = [[]]
          | k' >= n   = [l]
          | null l    = []
          | otherwise = map (y :) (combinations' (n - 1) (k' - 1) ys) ++ combinations' (n - 1) k' ys 


ghci> combinations 5 "abcdef"
["abcde","abcdf","abcef","abdef","acdef","bcdef"]

该问题是一个重复的排列，有人已经给出了答案。对于非重复排列，请使用Data.List中的permutations。