Question

我正在编写一个Haskell函数，该函数接收一个字符串并用＆＃39; _＆＃39;替换字符串中的任何重复字符。到目前为止，我有这个：

markDups :: [Char] -> [Char]
markDups = dupsHelp []
where dupsHelp c [] = c
      dupsHelp c (x:xs)
          | x `elem` c = dupsHelp c xs
          | otherwise = dupsHelp (c ++ [x]) xs

代码删除重复的字符。

示例：＆＃34; Hello World＆＃34; - ＆GT; ＆＃34; Helo Wrd＆＃34;

我如何修改此代码以获得＆＃34; Hel_o W_r_d＆＃34;？谢谢。

Answer 1

您无需存储已经看过的角色。您可以从字符串的其余部分中删除所有出现的字符：

markDups :: [Char] -> [Char]
markDups []       = []
markDups ('_':xs) = '_' : markDups xs
markDups (x  :xs) = x   : markDups (map mark xs)
  where
    mark y = if x == y then '_' else y

Answer 2

这很简单：

markDups :: [Char] -> [Char]
markDups = dupsHelp []
where dupsHelp c [] = c
      dupsHelp c (x:xs)
          | x `elem` c = dupsHelp (c ++ "_") xs
--                                ~~~~~~~~~~
          | otherwise = dupsHelp (c ++ [x]) xs

您可以放弃x并将其替换为x，而不是简单地放弃'_'。

你也可以用左折来表达：

import Data.List (foldl')

markDups = reverse . snd . foldl' go ([], [])
  where
  go (seen, acc) c =
    ( c : seen          -- Add character to seen duplicates for next step.
    , if c `elem` seen  -- If character is in duplicates:
      then '_' : acc    --   Add an underscore to result.
      else c : acc      --   Otherwise, just add the character.
    )

用＆＃39; _＆＃39;替换字符串中的重复字符char

2 个答案: