幼稚的递归斐波那契的复杂度确实为2?。
T(n) = T(n-1) + T(n-2) = T(n-2) + T(n-3) + T(n-3) + T(n-4) =
= T(n-3) + T(n-4) + T(n-4) + T(n-5) + T(n-4) + T(n-5) + T(n-5) + T(n-6) = ...
在每个步骤中,您调用T
两次,从而最终提供以下渐近性障碍:T(n) = 2?2?...?2 = 2?
Fib(n) = (φ? – (–φ)??)/sqrt(5) [where φ is the golden ratio]
(但是,由于浮点运算,它在现实生活中会遇到精度误差,这是不准确的)