Question

e.g。 lines "One string\non\neach line" = ["One string","on","each line"]

在列表，列表构造函数上使用模式匹配和递归，但没有其他函数。

这是我的解决方案，我从逻辑上思考，但它仍然给我一个错误。

lines :: [Char] -> [[Char]]
lines [] = [[]]
lines (x:'\n':xs) = [x] : lines xs
lines (x:xs) = x : lines xs

Answer 1

假设您已经知道lines除了输入的第一个字符之外的所有结果。你如何将第一个字符添加到该结果中？

charon :: Char -> [[Char]] -> [[Char]]
charon '\n'     css  = [] : css    -- to begin with newline, insert blank line
charon c         []  = [[c]]       -- very last char is in line of its own
charon c (cs : css)  = (c : cs) : css  -- else put char in first line

随着这个谜团的解决，

lines = foldr charon []

多年来，我一直让学生用拳头敲打家具，用 emp 吟唱 ty list ？使用 x cons xs ？＆＃34; 有时它会有所帮助。

Answer 2

首先，您需要一种在第一个换行符之前检索所有文本的方法。这是简单的递归：

beforeNewLine :: [Char] -> [Char]
beforeNewLine ""        = ""
beforeNewLine ('\n':xs) = ""
beforeNewLine (x:xs)    = x : beforeNewLine xs

同样，我们需要能够在第一个换行符之后获取所有文本：

afterNewLine :: [Char] -> [Char]
afterNewLine ""        = ""
afterNewLine ('\n':xs) = xs
afterNewLine (x:xs)    = afterNewLine xs

把它们放在一起：

myLines :: [Char] -> [[Char]]
myLines "" = []
myLines xs = (beforeNewLine xs) : myLines (afterNewLine xs)

实现haskell中的功能行

2 个答案: