递归地反转字符串（或列表）

发布于 2024-12-08 09:19:57 字数 540 浏览 0 评论 0原文

我正在尝试在 haskell 中编写一个函数来递归地反转列表。我编写了一个辅助函数，它采用原始列表和一个空列表，然后以 LIFO 模式将元素从第一个列表传输到另一个列表。

这就是我所拥有的：

myreverse :: [a] -> [a]
myreverse list = myflip list []

myflip :: [a] -> [a] -> [a]
myflip list1 newList
    | null list1        = newList
    | otherwise         = myflip (tail list1) ((head list1) : newList)

我知道有一个内置函数可以为我完成此操作，但要求是我只使用 head、tail、elem 和 null （也没有模式匹配）。所以我的问题是：是否有更好的解决方案，我只有一个函数 myreverse，仅消耗一个列表？（当然满足上述要求）

谢谢！

原文

I'm trying to write a function in haskell that reverses lists recursively. I wrote a helper function that takes the original list and an empty list then transfers elements from the first one to the other in a LIFO pattern.

Here's what I have:

myreverse :: [a] -> [a]
myreverse list = myflip list []

myflip :: [a] -> [a] -> [a]
myflip list1 newList
    | null list1        = newList
    | otherwise         = myflip (tail list1) ((head list1) : newList)

I know there's a built-in function that does it for me but the requirement is that I only use head, tail, elem and null (No pattern matching either). So my question is: is there a better solution where I only have one function, myreverse, that consumes only one list? (That meets the requirements above, of course)

Thanks!

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忆悲凉 2024-12-15 09:19:57

您可以尝试使用 foldl 反转列表，如下所示：

reverse' :: [a] -> [a]  
reverse' = foldl (\acc x -> x : acc) []

You can try reversing a list using foldl like so:

reverse' :: [a] -> [a]  
reverse' = foldl (\acc x -> x : acc) []

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烟织青萝梦 2024-12-15 09:19:57

因此，对于任何可能感兴趣的人，这就是我最终写的内容：

myreverse :: [a] -> [a]
myreverse list = myflip list []
    where myflip list newList = if null list then newList
                                else myflip (tail list) ((head list):newList)

感谢大家的评论和建议。

So for anyone who might be interested, this is what I ended up writing:

myreverse :: [a] -> [a]
myreverse list = myflip list []
    where myflip list newList = if null list then newList
                                else myflip (tail list) ((head list):newList)

Thanks everyone for your comments and suggestions.

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转身泪倾城 2024-12-15 09:19:57

将其保存在reverse.hs文件中

 reverseString :: [Char] -> [Char]
 reverseString [] = []
 reverseString (x:xs) = reverseString xs ++ [x]

然后运行
反向字符串“abc”

中国篮球协会

save this in reverse.hs file

 reverseString :: [Char] -> [Char]
 reverseString [] = []
 reverseString (x:xs) = reverseString xs ++ [x]

then run
reverseString "abc"

cba

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狼性发作 2024-12-15 09:19:57

除了满足您的额外要求所需的调整之外，您的功能相当于 reverse 在 GHC 中实现。因此我认为你的功能是相当不错的。

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爱你是孤单的心事 2024-12-15 09:19:57

这个功能满足您的要求，但效率非常低：

rev xs = if null xs then xs else rev (tail xs) ++ [head xs]

您的解决方案一点也不差。使用模式匹配并使 myFlip 成为本地函数后，它看起来不会难看。将局部函数与累加器一起使用的技术非常常见（尽管不是在如此简单的情况下）。

This function fulfills your requirements, but is horrible inefficient:

rev xs = if null xs then xs else rev (tail xs) ++ [head xs]

Your solution isn't bad at all. After using pattern matching and making myFlip a local function it wouldn't look ugly. And the technique of using a local function with an accumulator is very common (although not in such easy cases).

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