在haskell中反转列表 [英] Reverse a list in haskell

问题描述

以下是我的代码：

  reverseList :: [Int]  - > [Int] 
 reverseList [] = [] 
 reverseList（x：xs）= x：reverseList xs 
  

最终发生的事情是我最终以相同的顺序返回列表。我甚至有一个解决方案来解决这个问题，但我想知道我在这里做错了什么？我对哈斯克尔很新，所以我认为我应该更多地关注于理解，然后我可以轻松地解决更多的问题。我知道这个问题有很多解决方案，但是我需要更多的帮助来理解esp在这段代码中做错了什么。

解决方案

您将列表分隔为头部和尾部，但随后按照相同的顺序重新组装列表。以 [1，2，3] 为例：

在第一次调用中， x 将是 1 ，并且 xs 将会是 [2,3] 。然后你创建一个新的列表，在前面包含 x （so 1），后面跟着 reverseList [2，3 。

I am trying to reverse a list.

Following is my code:

reverseList :: [Int] -> [Int]
reverseList [] = []
reverseList (x:xs) =  x:reverseList xs

What ends up happening is I end up getting the list back in same order. I even have a solution for how to reverse the list but I am trying to understand what I did wrong here ? I am very new to haskell so think I should focus on understanding more then I can solve more problems with ease. I know there are many solutions out there for this problem but I need more help in the understanding esp what I did wrong here in this code.

解决方案

You are separating the list into head and tail, but then re-assemble the list in the same order. Take the list [1, 2, 3] for example:

In the first call, x will be 1, and xs will be [2, 3]. Then you create a new list, consisting of x (so 1) at the front, followed by reverseList [2, 3.

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