[英]How to sum elements of two lists. Haskell
我才剛剛開始學習Haskell,目前正在探索列表的可能性。 我想總結兩個列表,但不知何故出錯了。
所以:
輸入:sumTwoLists [2,5,7,7,9] [1,2,2](基本上是 25779 + 122)
Output:[2,5,9,0,1]
首先,我將整個列表顛倒過來,因為多位數字的添加必須從末尾開始:
reverseList :: [Int] -> [Int]
reverseList [] = []
reverseList (x:xs) = reverseList xs ++ [x]
有用。 然后我實現了一個add function:
add :: (Num a) => [a] -> [a] -> [a]
add _ [] = []
add [] _ = []
add (x:xs) (y:ys) = (x + y) : add xs ys
但是當一個列表比另一個列表短時,它就會出錯。 (add [2,5,7,7,9] [1,2,2] = [3,7,9]) 所以finction也必須在小數后面加0 ([1,2,2] = [1,2,2,0,0]。)
之后,我嘗試像這樣實現 sumTwoLists function:
sumTwoLists :: [Int] -> [Int] -> [Int]
sumTwoLists (x:xs) (y:ys) = reverseList ((reverseList (x:xs)) add (reverseList (y:ys)))
但是這段代碼沒有考慮元素不能大於9的事實。我不想將元素轉換為Int或Integer,這就是我不使用它們的原因。
我基本上只是想反轉列表,然后在最短列表中添加 0,然后將每個元素與另一個列表中的元素相加,如果結果 > 9,則結果除以 10(mod?)和相鄰數增加
如有任何幫助,我將不勝感激!
用列表來做這件事沒有多大意義,因為它會引入很多額外的問題:
9
並因此引入進位。 add
function 無法正常工作,因為它從列表之一用盡時停止,此外它沒有考慮到數字可能“溢出”。 因此,我們應該構造一個 function add
,它有一個額外的參數:進位:
add' :: Int -> [Int] -> [Int] -> [Int]
add' 0 [] [] = []
add' n [] [] = [n]
add' n xs [] = add' n xs [0]
add' n [] xs = add' n [0] xs
add' n (x:xs) (y:ys) = r : add' q xs ys
where (q,r) = quotRem (x+y+n) 10
因此add
以零作為進位開始:
add :: [Int] -> [Int] -> [Int]
add = add' 0
如果我們因此計算反向列表的總和,我們得到:
Prelude> add [9,7,7,5,2] [2,2,1]
[1,0,9,5,2]
為了使它工作,您需要使用add
作為中綴運算符,並且兩個操作數可以是空或非空列表:
sumTwoLists :: [Int] -> [Int] -> [Int]
sumTwoLists xs ys = reverseList ((reverseList xs) `add` (reverseList ys))
或使用on:: (b -> b -> c) -> (a -> b) -> a -> a -> c
:
import Data.Function(on)
sumTwoLists :: [Int] -> [Int] -> [Int]
sumTwoLists = add `on` reverse
對於給定的樣本輸入,我們現在得到:
Prelude> sumTwoLists [2,5,7,7,9] [1,2,2]
[2,5,9,0,1]
最后, reverseList
已經存在於Prelude
中: reverse:: [a] -> [a]
。 您實現的reverseList
function 在O(n 2 )中運行,效率不高,您可以使用累加器來獲得線性時間。
這是以 10 為底的進數和。這是一個實現。
-- odometer (add one in base b)
odom :: Int -> [Int] -> [Int]
odom b (x : xs) | x<(b-1) = (x+1) : xs
| xs==[] = [0,1]
| otherwise = 0 : odom b xs
-- iterated odometer
odom_iter :: Int -> [Int] -> Int -> [Int]
odom_iter b t n | n == 0 = t
| otherwise = odom b (odom_iter b t (n-1))
-- adic addition
sumadic :: Int -> [Int] -> [Int] -> [Int]
sumadic b (x:xs) (y:ys) | xs == [] = odom_iter b (y:ys) x
| ys == [] = odom_iter b (x:xs) y
| sumxy < b = (sumxy : (sumadic b xs ys))
| sumxy == b = (0 : (odom b (sumadic b xs ys)))
| otherwise = (1 : (odom b (sumadic b xs ys)))
where sumxy = x+y
這通過顛倒列表來工作:
> sumadic 10 [9, 7, 7, 5, 2] [2, 2, 1]
[1,0,9,5,2]
因此,您只需將sumadic
應用於反轉列表並反轉 output:
sumTwoLists x y = reverse $ sumadic 10 (reverse x) (reverse y)
這個答案將使用 Haskell 的內置reverse
而不是您的reverseList
,盡管它們是可以互換的。
讓我們先看看您添加的 function:
add :: (Num a) => [a] -> [a] -> [a]
-- look at these next 2 lines specfically
add _ [] = []
add [] _ = []
add (x:xs) (y:ys) = (x + y) : add xs ys
如果您將一個非空列表添加到一個空列表,您當前的代碼說它是一個空列表,這顯然不是真的( add [1, 2, 3] []
應該是[1, 2, 3]
)。 因此,您可以首先通過返回另一個列表來修復您的基本情況:
add :: (Num a) => [a] -> [a] -> [a]
-- return the other list
add [] x = x
add x [] = x
add (x:xs) (y:ys) = (x + y) : add xs ys
如果您要添加兩個空列表,則另一個列表是空列表,因此您仍然可以正確返回空列表。 現在,我們可以解決“這段代碼沒有考慮元素不能大於9”的部分。 由於您的加法方法是模擬您如何在筆和紙上進行加法,因此請繼續這樣做。 例如,如果結果為 12,則數字為 2,並且攜帶 1。 Remember that mod is the remainder of division, so 12 `mod` 10
( backticks in Haskell makes a function infix ) is 2
, and 12 `div` 10
, due to the nature of integer division rounding down, will give you every digit after第一個數字,即1
。
話不多說,讓我們編寫一些代碼:
-- change Num to Integral because we need to work with integers
add :: (Integral a) => [a] -> [a] -> a -> [a]
-- we need to add a carry now ^
-- these base cases break down if carry is non-zero
add [] x c
-- if carry is zero we're fine
| c == 0 = x
-- just add the carry in as a digit
| otherwise = add [c] x 0
-- same applies here
add x [] c
| c == 0 = x
| otherwise = add x [c] 0
add (x:xs) (y:ys) c = dig : add xs ys rst
where sum = x + y + c -- find the sum of the digits plus the carry
-- these two lines can also be written as (rst, dig) = sum `divMod` 10
dig = sum `mod` 10 -- get the last digit
rst = sum `div` 10 -- get the rest of the digits (the new carry)
現在,您的助手 function 可以使用 0 作為初始進位調用它:
addTwoLists :: (Num a) => [a] -> [a] -> [a]
addTwoLists x y = reverse $ add (reverse x) (reverse y) 0
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