haskell - 列表的排列 - Haskell

Question

我想用 2 个列表制作所有可能的子组组合。这是一个执行此操作的函数:

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

如果你将“abc”传递给这个函数，它会返回:

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

对相同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"]

使它扩展到任意数量的列表的最简单方法是什么？下面是类型声明的样子:

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

Answer 1

你要的是 replicateM :

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

定义很简单:

replicateM n = sequence . replicate n

所以它是 sequence在这里做真正工作的单子(monad)名单上。