Question

处理给定SuffixTree作为输入的函数，输出该后缀树中的整数列表。例如。 getIndices tree1 = [2,4,1,3,5,0]。整数列表的顺序无关紧要。我收到错误，在函数的第二行：“Couldn't match expected type 'SuffixTree' with actual type '[SuffixTree]'”。我已经考虑了很久了，没有运气。任何帮助将不胜感激。

data SuffixTree = Leaf Int | Node [ ( String, SuffixTree ) ] 
            deriving (Eq,Ord,Show)

text1 :: String
text1 = "banana"

tree1 :: SuffixTree
tree1 = Node [("banana",Leaf 0),
              ("a",Node [("",Leaf 5),
                         ("na",Node [("",Leaf 3),
                                     ("na",Leaf 1)])]),
              ("na",Node [("",Leaf 4),
                          ("na",Leaf 2)])]

------------------------------------------------------------------

getIndices :: SuffixTree -> [ Int ]
getIndices sufTree = getIndices' sufTree [] 
  where getIndices' :: SuffixTree -> [Int] -> [Int]
        getIndices' (Node ((_, Node xs):ys)) c 
          | Node xs == Node [] = c
          | otherwise = getIndices' ((Node xs):([Node ys])) c
        getIndices' (Node ((_,Leaf i):xs)) c = getIndices' (Node xs) (i:c)

Answer 1

您的getIndices'实用程序函数声明为SuffixTree，但在otherwise情况下，您传递(Node xs:[Node ys])类型为[SuffixTree]的{{1}}。

鉴于你要做的就是收集树中的整数，或许你的otherwise案例只需要调用getIndices'两次：

| otherwise = getIndices' (Node xs) (getIndices' (Node ys) c)

您的代码还有其他一些问题。如果使用警告（-Wall）进行编译，编译器将警告您模式匹配不完整。您的代码也因运行而在运行时失败。

不完整是因为getIndices'未涵盖所有可能的SuffixTree种类。您还需要填写getIndices' (Leaf Int)和getIndices' (Node [])的案例。

此外，| Node xs == Node []案例中Node ((_, Node xs):ys的现有案例会变得多余：它将通过getIndices'案例中对otherwise的递归调用来处理，那么新的Node []案例。您也可以考虑如何将已经存在的两种情况简化为单个案例。

Haskell：代数类型错误（后缀树：递归）

1 个答案: