将字符串列表合并到这些字符串的元组列表中答案

【问题标题】：Merge list of strings into list of tuple of those strings将字符串列表合并到这些字符串的元组列表中
【发布时间】：2017-08-29 03:13:00
【问题描述】：

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 中编程，一直在尝试解决这个问题。

【问题讨论】：

标签： string list haskell merge tuples

【解决方案1】：

import Data.List (inits, tails)

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

inits xs返回xs的所有前缀（包括[]和原始列表），tails xs返回xs的所有后缀（包括[]和原始列表）。由于包含了原始列表和[]，所以后面我们必须使用init和tail。

以下是一些中间结果：

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","")]

最后一行说明为什么zip 之后的init . tail 是必要的。

【讨论】：

【解决方案2】：

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]]

【讨论】：

如何将其放入函数中或调用它？不能是预定值
我不明白这个问题。 h 是你必须调用的函数。