将字符串列表合并到这些字符串的元组列表中

Question

我需要这样做：

mergers ["co","nt","ro","la"] ⇒ [("co","ntrola"),("cont","rola")("contro","la")]
mergers ["co","nt"] ⇒ [("co","nt")]
mergers ["co"] ⇒ []

但我不知道怎么做。我最近开始在Haskell编程，但我一直试图解决这个问题。

Answer 1

你可以试试这个，作为第一个解决方案：

a=["12","34","56", "78"]
g n x= (concat $ fst y, concat $ snd y) where y=splitAt n x
h a = [g i a|i<-[1..(length a)-1]]

Answer 2

import Data.List (inits, tails)

mergers :: [[a]] -> [([a], [a])]
mergers xs = init $ tail $ zip (map concat $ inits xs) (map concat $ tails xs)

inits xs会返回xs的所有前缀（包括[]和原始列表），tails xs会返回xs的所有后缀（包括{{1}和原始列表）。由于原始列表和[]已包含在内，我们之后必须使用[]和init。

以下是一些中间结果：

tail

最后一行显示了ghci> inits ["co","nt","ro","la"] [[],["co"],["co","nt"],["co","nt","ro"],["co","nt","ro","la"]] ghci> tails ["co","nt","ro","la"] [["co","nt","ro","la"],["nt","ro","la"],["ro","la"],["la"],[]] ghci> map concat $ inits ["co","nt","ro","la"] ["","co","cont","contro","controla"] ghci> map concat $ tails ["co","nt","ro","la"] ["controla","ntrola","rola","la",""] ghci> let xs = ["co","nt","ro","la"] in zip (map concat $ inits xs) (map concat $ tails xs) [("","controla"),("co","ntrola"),("cont","rola"),("contro","la"),("controla","")] 之后需要init . tail的原因。