entityMap|blocks|key|2qka3|text|您将列表分为头部和尾部，但随后以相同的顺序重新组合列表。拿着清单[1,+2,+3]例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|75odo|在第一个调用中，x将会是1，以及xs将会是[2,+3]..。然后创建一个新列表，该列表包含x(所以1)在前面，然后是reverseList+[2,+3]..。^0|W|9|0|8|1|C|1|G|2|L|6|19|1|1M|I^^$0|$]|1|@$2|3|4|5|6|7|8|I|9|@$A|J|B|K|C|D]]|E|@]|F|$]]|$2|G|4|H|6|7|8|L|9|@$A|M|B|N|C|D]|$A|O|B|P|C|D]|$A|Q|B|R|C|D]|$A|S|B|T|C|D]|$A|U|B|V|C|D]|$A|W|B|X|C|D]]|E|@]|F|$]]]]

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

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

entityMap|blocks|key|fg4r4|text|在Haskell中有几种方法可以解决这个问题。简单的方法是使用concatenate函数%2B%2B|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|cea5a|reverseList+[]+=+[]
reverseList+(x:xs)+=+reverseList+xs+%2B%2B+[x]|code-block|syntax|javascript|1ud9a|然而，对于大型列表来说，这会非常慢，因为Haskell列表实际上是单链表，所以为了附加一个元素，你必须遍历整个列表。另一种方法是跟上你在helper函数中构建的列表：|dba2a|reverseList+=+go+[]
++++where
++++++++go+acc+[]+=+acc
++++++++go+acc+(x:xs)+=+go+(x:acc)+xs|1vfhg|然而，这实际上只是fold模式：|1ae39|reverseList+=+foldl+(\acc+x+->+x+:+acc)+[]|js|6u4i|但是\acc+x+->+x+:+acc只是flip+(:)，所以这可以写成|fpubb|reverseList+=+foldl+(flip+(:))+[]|7e1o2|但是，最简单的方法可能是只使用reverse前言中的函数。|e4tav|我想指出的是，你的类型reverseList+::+[Int]+->+[Int]可以概括为::+[a]+->+[a]，你不需要对列表中的元素做任何特殊的事情，你只需要用它们构建一个新的列表即可。^0|18|2|0|0|0|0|9|4|0|0|2|H|L|8|0|0|F|7|0|B|T|19|D^^$0|$]|1|@$2|3|4|5|6|7|8|12|9|@$A|13|B|14|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|15|9|@]|E|@]|F|$J|K]]|$2|L|4|M|6|7|8|16|9|@]|E|@]|F|$]]|$2|N|4|O|6|I|8|17|9|@]|E|@]|F|$J|K]]|$2|P|4|Q|6|7|8|18|9|@$A|19|B|1A|C|D]]|E|@]|F|$]]|$2|R|4|S|6|I|8|1B|9|@]|E|@]|F|$J|T]]|$2|U|4|V|6|7|8|1C|9|@$A|1D|B|1E|C|D]|$A|1F|B|1G|C|D]]|E|@]|F|$]]|$2|W|4|X|6|I|8|1H|9|@]|E|@]|F|$J|T]]|$2|Y|4|Z|6|7|8|1I|9|@$A|1J|B|1K|C|D]]|E|@]|F|$]]|$2|10|4|11|6|7|8|1L|9|@$A|1M|B|1N|C|D]|$A|1O|B|1P|C|D]]|E|@]|F|$]]]]

There are several ways to solve this problem in Haskell. The naive approach would be to use the concatenate function <code>++</code>:

<pre><code>reverseList [] = []
reverseList (x:xs) = reverseList xs ++ [x]
</code></pre>

However, this will be really slow for large lists since Haskell lists are really singly linked lists, so in order to append an element you have to traverse the entire list. An alternative would be to keep up with the list you're building in a helper function:

<pre><code>reverseList = go []
 where
 go acc [] = acc
 go acc (x:xs) = go (x:acc) xs
</code></pre>

However, this is really just the <code>fold</code> pattern:

<pre><code>reverseList = foldl (\acc x -&gt; x : acc) []
</code></pre>

But <code>\acc x -&gt; x : acc</code> is just <code>flip (:)</code>, so this can be written as

<pre><code>reverseList = foldl (flip (:)) []
</code></pre>

However, the easiest way would probably be to just use the <code>reverse</code> function in Prelude.

I would like to point out that your type of <code>reverseList :: [Int] -&gt; [Int]</code> could be generalized to <code>:: [a] -&gt; [a]</code>, you don't do anything special with the elements of the list, you're just building a new list with them.

entityMap|blocks|key|4fq0p|text|在Haskell中有几种方法可以解决这个问题。下面是一个包含cons和last/init的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|bq2k5|reverseList++[]+=+[]
reverseList++xs+=+last+xs+:+reverseList+(init+xs)|code-block|syntax|javascript|3ienv|或者使用foldl：|7g5m5|reverseList+xs+=+foldl+(\x+y+->+y:x)+[]+xs|js^0|0|0|0^^$0|$]|1|@$2|3|4|5|6|7|8|M|9|@]|A|@]|B|$]]|$2|C|4|D|6|E|8|N|9|@]|A|@]|B|$F|G]]|$2|H|4|I|6|7|8|O|9|@]|A|@]|B|$]]|$2|J|4|K|6|E|8|P|9|@]|A|@]|B|$F|L]]]]

There are several ways to solve this problem in Haskell. Here a solution with cons and last/init: 

<pre><code>reverseList [] = []
reverseList xs = last xs : reverseList (init xs)
</code></pre>

Or with foldl:

<pre><code>reverseList xs = foldl (\x y -&gt; y:x) [] xs 
</code></pre>

entityMap|0|type|LINK|mutability|MUTABLE|data|url|http://hackage.haskell.org/package/dlist-0.8.0.6/docs/Data-DList.html|blocks|key|49qrc|text|基本上是使用追加的朴素算法|unstyled|depth|inlineStyleRanges|entityRanges|2urk7|revList+[]+=+[]
revList+(x:xs)+=+revList+xs+%2B%2B+[x]|code-block|syntax|javascript|37bah|是低效的，因为追加是一个O(n)操作n参数的第一个(左)参数的长度。%2B%2B运算符。所以revList上面的函数被证明是O(n(n-1)/2)+~+O(n%5E2)。|offset|length|style|CODE|24nn4|因此，对于这样的附加繁重任务，存在差异列表数据类型。|dfb1p|差异列表只是一个表示为函数的列表。我的意思是，像这样的列表[1,2,3]当用DList类型表示时，将是\xs+->+[1,2,3]+%2B%2B+xs或者简而言之([1,2,3]+%2B%2B)|a591f|type+DList+a+=+[a]+->+[a]

toDList+::+[a]+->+DList+a
toDList+xs++=+(xs+%2B%2B+)

fromDList+::+DList+a+->+[a]
fromDList+f+=+f+[]|551kc|这有点酷，因为因为DLists是函数，所以我们可以通过组合(.)来附加它们。运算符，并获取新的DList。换句话说，toDList+(xs+%2B%2B+ys)+==+(toDList+xs)+.+(toDList+ys)..。|ckagg|那么这有什么用呢？通过使用嵌套函数组合，我们可以用类似的方式反转列表revList函数，但它的成本要低得多。只有O(n)，因为每个函数组合都是O(1)。|73cfk|revList'+::+[a]+->+DList+a
revList'+[]+++++=+toDList+[]
revList'+(x:xs)+=+revList'+xs+.+toDList+[x]|1p96v|现在我们有了反转[a]在DList+a键入我们需要应用的所有内容fromDList|a5705|fastReverse+::+[a]+->+[a]
fastReverse+=+fromDList+.+revList'|5n6bb|差异列表数据类型并不像我上面显示的那么简单。它可以有Monoid、Functor和MonadT实例。有关这个有用的数据类型的更多信息，请查看Data.DList^0|0|0|C|4|I|1|Y|2|16|7|0|0|T|7|1F|K|25|C|0|0|1M|1D|0|Y|7|0|0|8|3|C|7|W|9|0|0|1Y|A|0^^$0|$1|$2|3|4|5|6|$7|8]]]|9|@$A|B|C|D|2|E|F|1B|G|@]|H|@]|6|$]]|$A|I|C|J|2|K|F|1C|G|@]|H|@]|6|$L|M]]|$A|N|C|O|2|E|F|1D|G|@$P|1E|Q|1F|R|S]|$P|1G|Q|1H|R|S]|$P|1I|Q|1J|R|S]|$P|1K|Q|1L|R|S]]|H|@]|6|$]]|$A|T|C|U|2|E|F|1M|G|@]|H|@]|6|$]]|$A|V|C|W|2|E|F|1N|G|@$P|1O|Q|1P|R|S]|$P|1Q|Q|1R|R|S]|$P|1S|Q|1T|R|S]]|H|@]|6|$]]|$A|X|C|Y|2|K|F|1U|G|@]|H|@]|6|$L|M]]|$A|Z|C|10|2|E|F|1V|G|@$P|1W|Q|1X|R|S]]|H|@]|6|$]]|$A|11|C|12|2|E|F|1Y|G|@$P|1Z|Q|20|R|S]]|H|@]|6|$]]|$A|13|C|14|2|K|F|21|G|@]|H|@]|6|$L|M]]|$A|15|C|16|2|E|F|22|G|@$P|23|Q|24|R|S]|$P|25|Q|26|R|S]|$P|27|Q|28|R|S]]|H|@]|6|$]]|$A|17|C|18|2|K|F|29|G|@]|H|@]|6|$L|M]]|$A|19|C|1A|2|E|F|2A|G|@]|H|@$P|2B|Q|2C|A|2D]]|6|$]]]]

Basically the naive algorithm which uses appending

<pre><code>revList [] = []
revList (x:xs) = revList xs ++ [x]
</code></pre>

is inefficient since appending is an <code>O(n)</code> operation where <code>n</code> is the length of the first (left) parameter of the <code>++</code> operator. So the <code>revList</code> function above turns out to be O(n(n-1)/2) ~ O(n^2).

So for such append heavy tasks there are the Difference List data type.

A difference list is just a list expressed as a function. What i mean is, a list like <code>[1,2,3]</code> when expressed in DList type would be <code>\xs -&gt; [1,2,3] ++ xs</code> or in short <code>([1,2,3] ++)</code>

<pre><code>type DList a = [a] -&gt; [a]

toDList :: [a] -&gt; DList a
toDList xs = (xs ++ )

fromDList :: DList a -&gt; [a]
fromDList f = f []
</code></pre>

This is sort of cool because since DLists are functions we can append them by composition (.) operator and get a new DList. In other words <code>toDList (xs ++ ys) == (toDList xs) . (toDList ys)</code>.

So how is this useful? By using nested function compositions we can reverse our list in a similar fashion to <code>revList</code> function but it will cost us much less. Only O(n) since every function composition is O(1).

<pre><code>revList' :: [a] -&gt; DList a
revList' [] = toDList []
revList' (x:xs) = revList' xs . toDList [x]
</code></pre>

Now that we have the reversed <code>[a]</code> in <code>DList a</code> type all we need to apply <code>fromDList</code>

<pre><code>fastReverse :: [a] -&gt; [a]
fastReverse = fromDList . revList'
</code></pre>

The Difference List data type is not as simple as i have shown above. It can have Monoid, Functor and MonadT instances. For more on this useful data type check <a href="http://hackage.haskell.org/package/dlist-0.8.0.6/docs/Data-DList.html" rel="nofollow noreferrer">Data.DList</a>

entityMap|blocks|key|3gndn|text|很简单。使用内置的reverse功能：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4peu3|print+(reverse+[1,2,3,4,5])+--+[5,4,3,2,1]|code-block|syntax|js^0|9|7|0^^$0|$]|1|@$2|3|4|5|6|7|8|L|9|@$A|M|B|N|C|D]]|E|@]|F|$]]|$2|G|4|H|6|I|8|O|9|@]|E|@]|F|$J|K]]]]

Simple. use the built-in <code>reverse</code> function:
<pre><code>print (reverse [1,2,3,4,5]) -- [5,4,3,2,1]
</code></pre>

I am trying to reverse a list.

Following is my code:

<pre><code>reverseList :: [Int] -&gt; [Int]
reverseList [] = []
reverseList (x:xs) = x:reverseList xs
</code></pre>

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.

Reverse a list in haskell

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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 我正在尝试颠倒一个列表。 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重新获得了列表。我甚至有一个如何反转列表的解决方案，但我正在试图...

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