algorithm - Erlang "case" clause for values in a range -

learning erlang, i'm solving simple problem , came solution:

%%%------------------------------------------------------------------ %%% @doc https://leetcode.com/problems/add-two-numbers/ %%% %%% @end %%%------------------------------------------------------------------ -module(add_two_numbers).  -define(base10, 10).  -export([main/2]).  -ifdef(test). -include_lib("eunit/include/eunit.hrl"). -endif.  -spec main(l1 :: nonempty_list(non_neg_integer()), l2 :: nonempty_list(non_neg_integer())) -> list().  %%%================================================================== %%% export %%%================================================================== main(l1, l2) ->   loop(l1, l2, 0, []).  -ifdef(test). main_test() ->   ?assertequal([0, 2, 1, 2, 1, 1], add_two_numbers:main([9, 6, 9, 5, 9], [1, 5, 1, 6, 1])),   ?assertequal([3, 6, 9, 7, 5], add_two_numbers:main([4, 1, 8, 7, 2], [9, 4, 1, 0, 3])),   ?assertequal([6, 7, 9, 0, 1], add_two_numbers:main([2, 2, 3, 3], [4, 5, 6, 7])),   ?assertequal([6, 3, 7, 4, 1], add_two_numbers:main([4, 1, 0, 8], [2, 2, 7, 6])),   ?assertequal([2, 7, 9, 1], add_two_numbers:main([2, 7, 1, 0], [0, 0, 8, 1])),   ?assertequal([8, 9, 9, 1], add_two_numbers:main([9, 9, 9], [9, 9, 9])),   ?assertequal([7, 1, 6, 1], add_two_numbers:main([9, 3, 7], [8, 7, 8])),   ?assertequal([3, 5, 6, 1], add_two_numbers:main([6, 6, 6], [7, 8, 9])),   ?assertequal([0, 0, 0], add_two_numbers:main([0, 0, 0], [0, 0, 0])),   ?assertequal([7, 0, 8], add_two_numbers:main([2, 4, 3], [5, 6, 4])),   ?assertequal([0, 2, 2], add_two_numbers:main([0, 1], [0, 1, 2])),   ?assertequal([0, 1, 1], add_two_numbers:main([4, 6], [6, 4])),   ?assertequal([0, 0, 1], add_two_numbers:main([1], [9, 9])),   ?assertequal([0, 1], add_two_numbers:main([], [0, 1])),   ?assertequal([], add_two_numbers:main([], [])),    ?asserterror(badarith, add_two_numbers:main([0, 1, 2], ["", 5, 6])). -endif.  %%%================================================================== %%% internal %%%================================================================== loop([h1 | t1], [h2 | t2], c, r) when h1 + h2 + c >= ?base10 ->   loop(t1, t2, 1, r ++ [h1 + h2 + c - ?base10]); loop([], [h | t], c, r) when h + c >= ?base10 ->   loop([], t, 1, r ++ [h + c - ?base10]); loop([h | t], [], c, r) when h + c >= ?base10 ->   loop([], t, 1, r ++ [h + c - ?base10]); loop([h1 | t1], [h2 | t2], c, r) ->   loop(t1, t2, 0, r ++ [h1 + h2 + c]); loop([], [h | t], c, r) ->   loop([], t, 0, r ++ [h + c]); loop([h | t], [], c, r) ->   loop([], t, 1, r ++ [h + c]); loop([], [], c, r) when c > 0 -> r ++ [c]; loop([], [], _, r) -> r. 

what bothers me how many loop calls have define. there might better solutions this.

my question: there way can tell in case-of if condition within range? this, instance:

x = h1 + h2 + c, case x of   x >= 10 -> blah;   _ -> ummm end. 

---

update

this trying achieve:

loop([h1 | t1], [h2 | t2], c, r) ->   case h1 + h2 + c >= ?base10 of     true -> loop(t1, t2, 1, r ++ [h1 + h2 + c - ?base10]);     false -> loop(t1, t2, 0, r ++ [h1 + h2 + c])   end; loop([], [h | t], c, r) ->   case h + c >= ?base10 of     true -> loop([], t, 1, r ++ [h + c - ?base10]);     false -> loop([], t, 0, r ++ [h + c])   end; loop([h | t], [], c, r) ->   case h + c >= ?base10 of     true -> loop(t, [], 1, r ++ [h + c - ?base10]);     false -> loop(t, [], 0, r ++ [h + c])   end; loop([], [], c, r) ->   case c > 0 of     true -> r ++ [c];     false -> r   end. 

...not sure if it's better though.

you use little trick

loop([], [], 0, r) -> lists:reverse(r); loop([], [], 1, r) -> lists:reverse(r, [1]); loop([], l2, c, r) ->     loop([0], l2, c, r); loop(l1, [], c, r) ->     loop(l1, [0], c, r); loop([h1 | t1], [h2 | t2], c, r) ->     case h1 + h2 + c of         s when s >= ?base10 ->             loop(t1, t2, 1, [s - ?base10 | r]);         s ->             loop(t1, t2, 0, [s | r])     end. 

note don't use acc++[x] pattern because bad habit make o(n^2)(see[1]). performance point of view tail call optimization not necessary if have use lists:reverse/1,2 non-tail call version should fast tail call optimised or on platforms faster:

main(l1, l2) ->   loop(l1, l2, 0).  loop([], [], 0) -> []; loop([], [], 1) -> [1]; loop([], l2, c) ->     loop([0], l2, c); loop(l1, [], c) ->     loop(l1, [0], c); loop([h1 | t1], [h2 | t2], c) ->     case h1 + h2 + c of         x when x >= ?base10 ->             [ x - ?base10 | loop(t1, t2, 1) ];         x ->             [ x | loop(t1, t2, 0) ]     end. 

or can make 1 step further , remove case , + 0 out

loop([], [], 0) -> []; loop([], [], 1) -> [1]; loop([], [h | t], c) ->     loop_([], t, h + c); loop([h | t], [], c) ->     loop_([], t, h + c); loop([h1 | t1], [h2 | t2], c) ->     loop_(t1, t2, h1 + h2 + c).  loop_(l1, l2, s) when s >= ?base10 ->     [ s - ?base10 | loop(l1, l2, 1) ]; loop_(l1, l2, s) ->     [ s | loop(l1, l2, 0) ]. 

[1]: try [l1, l2] = [ [rand:uniform(10)-1 || _ <- lists:seq(1, 100000)] || _ <- [1,2] ], timer:tc(fun() -> add_two_numbers:main(l1, l2) end). code. in code takes 3.5ms , yours 33s.