实现尾递归
Implementing tail recursion
我在 haskell 中编写了一个简单的函数,它是非尾递归的,它对列表中的值求和,其中:
nonTailRecursiveSum :: [Integer] -> Integer
nonTailRecursiveSum [] = 0 --base case
nonTailRecursiveSum (x:xs) = x + sum xs
但我现在要做的是实现相同的功能,但使用尾递归。据我所知,尾递归在最后一步执行递归调用所以我尝试了类似的东西:
tailRecursiveSum :: [Integer] -> Integer
tailRecursiveSum [] = 0
tailRecursiveSum (x:xs) = aux_f(x) + tailRecursiveSum xs
.
.
但是我在中途迷路了,因为我不熟悉 Haskell 中的尾递归。谁能帮我继续尾递归版本的代码?
玩一下,
sum (x:y:xs) = x + sum (y:xs)
= x + (y + sum xs)
= (x + y) + sum xs
g a b = a + sum b
sum (x:y:xs) = g x (y:xs)
= x + g y xs
= g (x+y) xs -- !!!
最后一个是尾递归形式!因此我们只定义
sum xs = g 0 xs
where
g acc [] = ...
g acc (x:xs) = g (acc + ...) ...
填空!
我在 haskell 中编写了一个简单的函数,它是非尾递归的,它对列表中的值求和,其中:
nonTailRecursiveSum :: [Integer] -> Integer
nonTailRecursiveSum [] = 0 --base case
nonTailRecursiveSum (x:xs) = x + sum xs
但我现在要做的是实现相同的功能,但使用尾递归。据我所知,尾递归在最后一步执行递归调用所以我尝试了类似的东西:
tailRecursiveSum :: [Integer] -> Integer
tailRecursiveSum [] = 0
tailRecursiveSum (x:xs) = aux_f(x) + tailRecursiveSum xs
.
.
但是我在中途迷路了,因为我不熟悉 Haskell 中的尾递归。谁能帮我继续尾递归版本的代码?
玩一下,
sum (x:y:xs) = x + sum (y:xs)
= x + (y + sum xs)
= (x + y) + sum xs
g a b = a + sum b
sum (x:y:xs) = g x (y:xs)
= x + g y xs
= g (x+y) xs -- !!!
最后一个是尾递归形式!因此我们只定义
sum xs = g 0 xs
where
g acc [] = ...
g acc (x:xs) = g (acc + ...) ...
填空!