Ninety-nine Lisp Problem #p04,#p05

P04 (*) Find the number of elements of a list.

ทำได้ 2 แบบคือ
1. ใช้ recursive ธรรมดา
2. ใช้ tail recursive

clisp

(defun len (lst)
  (if (null lst)
      0
    (1+ (len (cdr lst)))))

(defun len-2 (lst &optional (cnt 0))
  (if (null lst)
      cnt
    (len-2 (cdr lst) (1+ cnt))))

haskell

len :: [a] -> Int
len ([]) = 0
len (_:xs) = 1 + len xs


len_2 :: [a] -> Int
len_2 (xs) = len_helper xs 0
  where 
    len_helper ([]) n = n
    len_helper (_:xs) n = len_helper xs (n + 1)

erlang

-module(p04).
-export([len/1,len_2/1]).

len([]) ->
    0;
len([_|XS]) ->
    1 + len(XS).


len_2(L) ->
    len_iter(L,0).

len_iter([],N) ->
    N;
len_iter([_|XS],N) ->
    len_iter(XS,N+1).

P05 (*) Reverse a list.

implement ทั้งหมดแบบ tail-recursive

clisp

(defun rev (lst &optional (tmp '()))
  (if (null lst)
      tmp
    (rev (cdr lst) (cons (first lst) tmp))))

haskell

inv :: [a] -> [a]
inv l = invx l []
         where 
           invx [] y     = y
           invx (x:xs) y = invx xs (x : y)

ดูยาว, แต่ถ้าไปดู implement ของ hugs เขาจะเขียนแค่นี้

reverse   :: [a] -> [a]
reverse    = foldl (flip (:)) []

erlang

-module(p05).
-export([rev/1]).

rev(X) ->
    rev(X,[]).

rev([], L) ->
    L;
rev([H|T], L) ->
    rev(T, [H|L]).