带有两个合并函数asc和desc的Haskell合并排序

我想使用合并排序算法。mergeSort是等待合并函数作为第一个参数的主函数。有人知道我的问题在哪里吗？事先非常感谢

mergeSort xs = merge xs
mergeDesc xs = reverse (mergeAsc xs)
mergeAsc [] = []
mergeAsc [x] = [x]
mergeAsc xs = merge (mergeAsc top) (mergeAsc bottom) where (top, bottom) = splitAt (length xs `div` 2) xs
merge [] ys = ys
merge xs [] = xs
merge (x:xs) (y:ys) | x <= y    = x : merge xs (y:ys)
                    | otherwise = y : merge (x:xs) ys

---

将类型签名添加到函数中，问题就会变得很明显：

mergeAsc, mergeDesc :: Ord a => [a] -> [a]

mergeDesc xs = reverse (mergeAsc xs)
mergeAsc [] = []
mergeAsc [x] = [x]
mergeAsc xs = merge (mergeAsc top) (mergeAsc bottom) where (top, bottom) = splitAt (length xs `div` 2) xs

merge :: Ord a => [a] -> [a] -> [a]
merge [] ys = ys
merge xs [] = xs
merge (x:xs) (y:ys) | x <= y    = x : merge xs (y:ys)
                    | otherwise = y : merge (x:xs) ys

或者更简单更喜欢

mergeSort = margeAsc

请注意，mergeDesc并不是很好：您首先按错误的顺序对列表排序，然后将其反转？在Haskell中，您希望您的算法足够灵活，能够自行处理不同的排序之类的内容。所以你会定义

mergeSortBy :: (a->a->Ordering) -> [a] -> [a]
mergeSortBy cmp = mSort
 where 
       mSort [] = []
       mSort [x] = [x]
       mSort xs = merge (mSort top) (mSort bottom)
        where (top, bottom) = splitAt (length xs `quot` 2) xs

       merge [] ys = ys
       merge xs [] = xs
       merge (x:xs) (y:ys) = case x`cmp`y of
          LT  -> x : merge xs (y:ys)
          _   -> y : merge (x:xs) ys

然后，您可以简单地定义mergeSort=mergeSortBy compare和mergeSortDesc=mergeSortBy flip compare

还要观察将merge设置为本地函数如何防止实现中的错误

它说它应该声明为：mergeSort:：[a]->[a]->[a]->[a]->[a]->[a]作为接受合并函数作为第一个参数的函数

这很奇怪，它不应该被称为mergeSort，而是sortWithMerge之类的。无论如何，这样做很简单：只需抛出仅在merge子函数中使用的cmp！并将其替换为merge作为参数，而不是在本地定义它

sortWithMerge :: ([a]->[a]->[a]) -> [a] -> [a]
sortWithMerge merger = mSort
 where 
       mSort [] = []
       mSort [x] = [x]
       mSort xs = merger (mSort top) (mSort bottom)
        where (top, bottom) = splitAt (length xs `quot` 2) xs

---

将类型签名添加到函数中，问题就会变得很明显：

mergeAsc, mergeDesc :: Ord a => [a] -> [a]

mergeDesc xs = reverse (mergeAsc xs)
mergeAsc [] = []
mergeAsc [x] = [x]
mergeAsc xs = merge (mergeAsc top) (mergeAsc bottom) where (top, bottom) = splitAt (length xs `div` 2) xs

merge :: Ord a => [a] -> [a] -> [a]
merge [] ys = ys
merge xs [] = xs
merge (x:xs) (y:ys) | x <= y    = x : merge xs (y:ys)
                    | otherwise = y : merge (x:xs) ys

或者更简单更喜欢

mergeSort = margeAsc

请注意，mergeDesc并不是很好：您首先按错误的顺序对列表排序，然后将其反转？在Haskell中，您希望您的算法足够灵活，能够自行处理不同的排序之类的内容。所以你会定义

mergeSortBy :: (a->a->Ordering) -> [a] -> [a]
mergeSortBy cmp = mSort
 where 
       mSort [] = []
       mSort [x] = [x]
       mSort xs = merge (mSort top) (mSort bottom)
        where (top, bottom) = splitAt (length xs `quot` 2) xs

       merge [] ys = ys
       merge xs [] = xs
       merge (x:xs) (y:ys) = case x`cmp`y of
          LT  -> x : merge xs (y:ys)
          _   -> y : merge (x:xs) ys

然后，您可以简单地定义mergeSort=mergeSortBy compare和mergeSortDesc=mergeSortBy flip compare

还要观察将merge设置为本地函数如何防止实现中的错误

它说它应该声明为：mergeSort:：[a]->[a]->[a]->[a]->[a]->[a]作为接受合并函数作为第一个参数的函数

这很奇怪，它不应该被称为mergeSort，而是sortWithMerge之类的。无论如何，这样做很简单：只需抛出仅在merge子函数中使用的cmp！并将其替换为merge作为参数，而不是在本地定义它

sortWithMerge :: ([a]->[a]->[a]) -> [a] -> [a]
sortWithMerge merger = mSort
 where 
       mSort [] = []
       mSort [x] = [x]
       mSort xs = merger (mSort top) (mSort bottom)
        where (top, bottom) = splitAt (length xs `quot` 2) xs

---

非常感谢您，我需要知道的最后一件事是，当我将mergeSort函数的类型签名设置为：mergeSort:：[a]->[a]->[a]->[a]->[a]->[a]时，应用程序中出现了类型错误。。。如何解决这个问题？这两个类型声明必须是您和我定义它的方式。到目前为止，谢谢@leftaroundabout:为什么您认为它应该是[a]->[a]->[a]->[a]->[a]->[a]？因为我有一个特定的任务要完成，它说它应该声明为：mergeSort:：[a]->[a]->[a]->[a]->[a]->[a]作为接受合并函数作为第一个参数的函数。。。还有两个单独的合并函数，它们对值进行升序和降序排序，如我的解决方案所示，这两个函数应该声明为：Ord a=>[a]->[a]任务说明：编写一个mergeSort合并排序函数，它等待合并函数作为第一个参数。mergeSort:：[a]->[a]->[a]->[a]->[a]->[a]写两个额外的合并函数，它们将对值进行升序/降序mergeDesc:：Ord a=>[a]->[a]->[a]mergeAsc:：Ord a=>[a]->[a]->[a]@baron ok。。。虽然正如我所说，这不应该被称为合并排序。至于mergeAsc和mergeDesc，我会再次将它们作为普通mergeGen:：a->a->Ordering->[a]->[a]->[a]->[a]的专业化实现，作为mergeAsc=mergeGen compare和mergeDesc=mergeGen$flip compare。非常感谢，这是我需要知道的最后一件事，当我将mergeSort函数的类型签名设置为：mergeSort:：[a]->[a]->[a]->[a]->[a]->[a]时，这是应用程序中的类型错误。。。如何解决这个问题？这两个类型声明必须是您和我定义它的方式。到目前为止，谢谢@leftaroundabout:为什么您认为它应该是[a]->[a]->[a]->[a]->[a]->[a]？因为我有一个特定的任务要完成，它说它应该声明为：mergeSort:：[a]->[a]->[a]->[a]->[a]->[a]作为接受合并函数作为第一个参数的函数。。。还有两个单独的合并函数，它们对值进行升序和降序排序，如我的解决方案所示，这两个函数应该声明为：Ord a=>[a]->[a]任务说明：编写一个mergeSort合并排序函数，它等待合并函数作为第一个参数。mergeSort:：[a]->[a]->[a]->[a]->[a]->[a]写两个额外的合并函数，它们将对值进行升序/降序mergeDesc:：Ord a=>[a]->[a]->[a]mergeAsc:：Ord a=>[a]->[a]->[a]@baron ok。。。虽然正如我所说，这不应该被称为合并排序。至于mergeAsc和mergeDesc，我将再次将它们作为通用mergeGen:：a->a->Ordering->[a]->[a]->[a]->[a]的专门化来实现 s mergeAsc=mergeGen比较和mergeDesc=mergeGen$flip比较。