问题描述
我试图使用递归树找出斐波那契数列的复杂度，并得出树的高度 = O(n) 最坏情况，每个级别的成本 = cn，因此complexity = n*n=n^2
I am trying to find complexity of Fibonacci series using a recursion tree and concluded height of tree = O(n) worst case, cost of each level = cn, hence complexity = n*n=n^2
怎么是O(2^n)?
推荐答案
一个朴素的递归斐波那契数列的复杂度确实是 2ⁿ.
The complexity of a naive recursive fibonacci is indeed 2ⁿ.
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，从而提供最终的渐近障碍:T(n) = 2⋅2⋅...⋅2 = 2ⁿ
In each step you call T twice, thus will provide eventual asymptotic barrier of:T(n) = 2⋅2⋅...⋅2 = 2ⁿ
奖励:斐波那契最好的理论实现实际上是一个闭合公式，使用黄金比例:
bonus: The best theoretical implementation to fibonacci is actually a close formula, using the golden ratio:
Fib(n) = (φⁿ – (–φ)⁻ⁿ)/sqrt(5) [where φ is the golden ratio]
(然而，由于浮点运算，它在现实生活中存在精度误差，不精确)
(However, it suffers from precision errors in real life due to floating point arithmetics, which are not exact)
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