从Haskell列表中删除重复项 [英] Remove duplicates from Haskell list

问题描述

我试图构建一个函数，它接受一个字符串的第一个元素，并从字符串中删除所有与它相同的其他元素。然后为第二个字符做同样的事。

即 - Heello将变成Helo和巧克力Chlate

我原来的尝试

  removeSuccessor :: String  - >字符串
 removeSuccessor x = [c | c <-x，x！ `elem` c] 
  

但是，这似乎不起作用..建议？

解决方案

您可以保留所有元素的集合，只保留当前元素b
$ b

  import Data.Set 
 removeDups :: Ord a => [a]  - >设置 - > [a] 
 removeDups [] sofar = [] 
 removeDups（x：rest）sofar 
 | member x sofar =（removeDups rest sofar）
 |用法：
 $ b 
 
 $ b 
  removeDupsHeello空 - Helo
 removeDups巧克力空 - 巧克力
  
我想，运行时间是 O（n log n）。  
 
 
或者您可以使用  nub   from  Data.List ：
  Prelude Data.List> import Data.List 
 Prelude Data.List> nubHeello
Helo
 Prelude Data.List> nubChocolate
Choclate
  
运行时间 O（n ^ 2）。
 
I am trying to build a function that takes the first element of a string, and removes all other elements equal to it from the string. Then does the same for the second character.

Ie - "Heello" would become "Helo" and "Chocolate" "Chlate"

My original attempt
removeSuccessor :: String -> String
removeSuccessor x = [c | c <- x, x ! `elem` c]
But that doesn't seem to work.. suggestions?
解决方案
You could keep a set of all elements seen and only keep the current one if it hasn't been seen yet:
import Data.Set
removeDups :: Ord a => [a] -> Set a -> [a]
removeDups [] sofar = []
removeDups (x:rest) sofar
     | member x sofar = (removeDups rest sofar)
     | otherwise      = x:(removeDups rest (insert x sofar))
Usage:
removeDups "Heello" empty    -- "Helo"
removeDups "Chocolate" empty -- "Choclate"
Run time is O(n log n), I think. 

Or you can use nub from Data.List:
Prelude Data.List> import Data.List
Prelude Data.List> nub "Heello"
"Helo"
Prelude Data.List> nub "Chocolate"
"Choclate"
Run-time is O(n^2).

                        
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