List Haskell列表理解-所有列表拆分的列表

我只想编写一个函数

splits

，它接受一个列表

l

，并返回一个元组列表，其中包含所有可能的拆分方法

l

所以它应该是这样工作的：

splits "Hello"
[("","Hello"),("H","ello"),("He","llo"),("Hel","lo"),("Hell","o"),("Hello","")]

实施1我写的是：

splits l = [(x,y) | i <- [0 ..length l], x <- take i l, y <- drop i l]

实施2正确的解决方案是

splits l = [(take i l, drop i l) | i <- [0 ..length l]]

---

splits l=[（take i l，drop i l）| i关键观察点是语句的内容
x啊，现在我明白了。我忘了
take i l
是一个列表和
x
splits l = [(take i l, drop i l) | i <- [0 ..length l]]

[i | i <-[1..3]] => [1,2,3]

[c | c <- "Word"] => "Word" or equivalently ['W','o','r','d']

splits l = [(x,y) | i <- [0 ..length l], x <- [take i l], y <- [drop i l]]

splits :: [a] -> [([a],[a])]
splits xx = splits' [] ([],xx)
  where splits' :: [([a],[a])]-> ([a],[a]) -> [([a],[a])]
        splits' acc xs@(_,[]) = reverse (xs:acc)
        splits' acc (xs,y:ys) = let xs' = (xs++[y],ys)
                                in splits' (xs':acc) xs'

splits :: [a] -> [([a],[a])]
splits xx = zipWith splitAt [0..(length xx)] (repeat xx)