1

下午好，

我是 Haskell 的新手，我正在尝试调整我制作的 QuickSort 算法，以使用元组列表对“Int”列表进行排序，但我完全不知道如何将尾部绑定到“a”让它按照我的需要工作，或者是否可以重用代码。这是我用来对“Int”列表进行排序的内容：

quickSort :: Ord a => [a] -> [a]
quickSort [] = []  
quickSort (x:xs) =   
let smallSort  = quickSort [a | a <- xs, a < x]  
    biggerSort = quickSort [a | a <- xs, a > x]  
in  smallSort ++ [x] ++ biggerSort

这是我试图用它做的，所以我可以对元组列表（Int，[Int]）进行排序。我想按元组的第一个元素对元组进行排序，所以如果我得到一个类似 [(2,[1]),(1,[]),(3,[2,1])] 的列表，它会返回这个[(1,[]),(2,[1]), (3,[2,1])]。

quickSortTuplas ((x,(ks)) : []) = [(x,(ks))]
quickSortTuplas ((x,(ks)):ps) =   
let smallSort  = quickSort [a | a <- ps, a < x]  
    biggerSort = quickSort [a | a <- ps, a > x]  
in  smallSort ++ [(x,(ks))] ++ biggerSort

如果我尝试加载它，我会收到以下错误：

 Occurs check: cannot construct the infinite type: a ~ (a, [a1])
* In the second argument of `(>)', namely `x'
  In the expression: a > x
  In a stmt of a list comprehension: a > x
* Relevant bindings include
    a :: (a, [a1]) (bound at reNuevoHaskell.hs:60:37)
    biggerSort :: [(a, [a1])] (bound at reNuevoHaskell.hs:60:9)
    ps :: [(a, [a1])] (bound at reNuevoHaskell.hs:58:27)
    ks :: [a1] (bound at reNuevoHaskell.hs:58:22)
    x :: a (bound at reNuevoHaskell.hs:58:19)
    quickSortTuplas :: [(a, [a1])] -> [(a, [a1])]
      (bound at reNuevoHaskell.hs:57:1)

非常感谢您能给我的任何见解。

4

1 回答 1

2

注意表达式中[a | a <- ps, a < x] 。a是一个元组，而x是一个Int. 因此a < x没有任何意义。无论如何，因为您的quicksort作品Ord a，您也可以使用它来订购元组列表。试试看`！

quickSort :: Ord a => [a] -> [a]
quickSort [] = []  
quickSort (x:xs) =   
  let smallSort  = quickSort [a | a <- xs, a < x]  
      biggerSort = quickSort [a | a <- xs, a > x]  
   in smallSort ++ [x] ++ biggerSort

main = print $ quickSort [(1,[2,3,4]) , (0, [4,5,6]), (2,[1,2,3])] -- This works fine
于 2021-05-19T11:26:29.770 回答