Question

我想在Haskell中使用这两个函数实现一个sort函数：

smallest :: (Ord a) => [a] -> a
smallest [] = error "empty list"
smallest [x] = x
smallest (x:xs)
  | x < smallest xs = x
  | otherwise = smallest xs

insert :: Int -> [Int] -> [Int]
insert x [] = [x]
insert x (y:ys)
  | x <= y = x:y:ys
  | otherwise = y:insert x ys

我的想法是使用递归在正确的位置插入最小的值但是因为我是Haskell的新手，所以我遇到了一些如何实现它的问题。

Answer 1

smallest (x:xs)
  | x < smallest xs = x
  | otherwise = smallest xs

重复列表中每个点的smallest个查询数量，呈指数级增长。代替：

smallest (x:xs) = min x (smallest xs)

，甚至只是smallest = minimum。以下是我可以通过您的功能或类似功能看到的几种类型：

insertionSort [] = []
insertionSort (x:xs) = insert x (insertionSort xs)

这个还需要smallest来回馈剩下的清单：

selectSmallest :: [Int] -> (Int, [Int])
selectSmallest (x:xs) = let (y, ys) = smallest xs in if x < y
  then (x, xs)
  else (y, x:ys)

selectionSorta [] = []
selectionSorta xs = let (y, ys) = smallest xs in
  y : selectionSorta ys

Haskell递归排序函数

1 个答案: