bendevski
Recursion using only original list
I’m playing around with elixir and solving some leetcode problems but I ran into an issue. The following code:
defmodule Solution do
@spec climb_stairs(n :: integer) :: integer
def climb_stairs(n) do
climb_stairs(n, [1,0])
end
def climb_stairs(1, lis) do
[first, second | _] = lis
first + second
end
def climb_stairs(n, lis) do
[first, second ] = lis
climb_stairs(n-1, [first + second, second])
end
end
always has lis be [1,0] (the result is always 1). Could someone explain what I’m doing wrong?
Marked As Solved
LostKobrakai
You never actually change the list:
lis = [1, 0] # this is what you start with
# hitting the first clause of `climb_stairs/2` you do:
[1, 0 | _] = lis
1 + 0
# returns 1
# hitting the second clause of `climb_stairs/2` you do:
[1, 0] = lis
climb_stairs(…, [1 + 0, 0])
# calls `climb_stairs` again with `[1, 0]`
# eventually this reaches the first clause getting to the above case.
Also Liked
benwilson512
Hi @bendevski welcome! It would be helpful if you could explain what this function is supposed to do.
Last Post!
bendevski
Thank you for the clear explanation. Mixed up the variables in my head and thought it was a language thing.
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