Question

嗨我正在尝试做一个函数输入是一个字符串列表，输出再次是输入中出现的所有单词的列表。

例如输入[“例如，”，“爱”，“讨厌。”] 输出[“For”，“example”，“love”，“hate”]

我有这个。任何帮助，将不胜感激。另外，我如何只用一个函数和线性时间删除空格？而不是使用任何现有的功能

split' :: String -> [String]
split' [] = []
split' (x:xs)
     | isBlank x = split' xs
     | otherwise = waitForBlank (x:xs) : split' (drop (length (waitForBlank (x:xs))) xs)

isBlank :: Char -> Bool
isBlank x = if x == ' ' then True else False

waitForBlank :: String -> String
waitForBlank [] = []
waitForBlank (x:xs)
     | isBlank x = []
     | otherwise = x : waitForBlank xs

Answer 1

有一个很酷的单行来执行你需要的东西

["For example,", "love,", "hate."] >>= words

>>=具有类型(>>=) :: Monad m => m a -> (a -> m b) -> m b，它接受一个返回monadic结构的函数，并将结果连接到monadic结构中。

如果您想自己实施words：

words' xs =
  let
    waitForBlank (acc, buff) [] = (acc ++ [buff], buff)
    waitForBlank (acc, buff) (x:xs) =
      if x == ' ' then
        waitForBlank (acc ++ [buff], []) xs
      else
        waitForBlank (acc, buff ++ [x]) xs
  in
    fst (waitForBlank ([], []) xs)

或者使用(:)和reverse结果（为了获得更好的效果）：

words'' xs =
  let
    waitForBlank (acc, buff) [] = (reverse (buff : acc), buff)
    waitForBlank (acc, buff) (x:xs) =
      if x == ' ' then
        waitForBlank ((reverse buff) : acc, []) xs
      else
        waitForBlank (acc, x:buff) xs
  in
    fst (waitForBlank ([], []) xs)

在haskell编程中拆分字符串

1 个答案: