初学者haskell程序员

Question

isPalindrome::(Eq a) => [a] -> Bool
isPalindrome [] = True
isPalindrome [x] = True
isPalindrome (x1:xs:x2:[]) 
    | x1 == x2 = isPalindrome xs
    |otherwise = False


[1 of 1] Compiling Main             ( myHas.hs, interpreted )

myHas.hs:37:27:
    Couldn't match expected type `[a]' against inferred type `a1'
      `a1' is a rigid type variable bound by
           the type signature for `isPalindrome' at myHas.hs:33:18
    In the first argument of `isPalindrome', namely `xs'
    In the expression: isPalindrome xs
    In the definition of `isPalindrome':
        isPalindrome (x1 : xs : x2 : [])
                       | x1 == x2 = isPalindrome xs
                       | otherwise = False

我是初学者haskell程序员,并不知道为什么我收到这个错误,请帮忙吗？

Answer 1

您将其视为xs列表,但(x1:xs:x2:[])假设它是输入列表的元素.

请注意,(x1:xs:x2:[])它只匹配包含3个元素的列表x1,xs并且x2将是类型的元素a.

所以xs是类型的a,但是当你将它传递给isPalindrome,我们只能假设它必须是东西的清单,这样类型的系统调用类型[a1].

编码所需内容的最简单方法是:

isPalindrome::(Eq a) => [a] -> Bool
isPalindrome l = l == (reverse l)

Answer 2

这是一个易于理解的答案,类似于您的尝试:

isPalindrome [] = True
isPalindrome [x] = True
isPalindrome xs = (head xs == last xs) && isPalindrome (init (tail xs))

因此,空元素或单元素列表是回文序列,如果第一个和最后一个元素相等,则较长列表是回文,而"中间"元素也是回文.

请注意,MGwynne的答案要高得多,因为上面的解决方案必须在每个步骤中遍历列表.

Answer 3

我觉得这里需要对列表使用的语法进行解释,到目前为止还没有给出.首先,Haskell中列表类型的定义是:

data [a] = a : [a] | []

其中列表是空的([])或者是由(:)构造函数构成的,它的左参数为a a,另一个列表([a]在定义中).这意味着列表[1,2,3]实际上是1 : (2 : (3 : [])),但这也可以写成1 : 2 : 3 : [].

当列表上的模式匹配时,您将匹配这些构造函数:

f [] = … -- match the empty list

f (x:[]) = … -- match a list with one element, which you name x

f (x:xs) = … -- match the first element of the list, and whatever the rest of 
             -- the list is, but it must have at least one element. if you call
             -- f [1,2,3], x will be bound to 1, and xs will be bound to [2,3]
             -- because [1,2,3] is the same as (1:[2,3])

f (x:y:xs) = … -- matches a list with at least two elements, which you 
               -- call x and y respectively

f (xs:ys:zs:things) = … -- matches a list with at least three elements,
                        -- which you name, xs, ys and zs.

因此,希望现在很清楚

f (x1:xs:x2:[])

匹配具有三个元素的列表,您将其命名为x1,xs和x2.

我希望有所帮助.