如何计算递归函数的上限时间复杂度（“大 O”）？
[英]How to calculate the upper bound time complexity (`“big O`”) of a recursive function?
问题描述
Suppose I have a recursive function T , and I want to calculate the upper bound timer complexity of this function.
假设我有一个递归函数T ，我想计算这个函数的上限计时器复杂度。
T(1) = 3 T(1) = 3
T(n) = 3T(n/3) + 3. T(n) = 3T(n/3) + 3。
How can I find the upper bound of the time complexity of T(n)?如何找到 T(n) 的时间复杂度的上限？
2 个解决方案
解决方案1
3 已采纳 2013-11-29 05:08:00
Use the master theorem case使用主定理案例where a=3, b=3, c=0.
其中 a=3，b=3，c=0。
I highly recommend the MIT lectures on Algorithms.我强烈推荐麻省理工学院关于算法的讲座。 You can learn more about the Master theorem in lecture 2您可以在第2 课中了解有关 Master 定理的更多信息
解决方案2
0 2013-11-29 03:59:07
assume that, n = 3^k假设，n = 3^k
F(0) = 3 F(0) = 3
F(k) = 3 * F(k-1) + 3 F(k) = 3 * F(k-1) + 3
= 3^2 * F(k-2) + 3^2 + 3
= ...
= 3^k * F(0) + 3^k + 3^(k-1) + ... + 3
= 3^(k+1) + 3^k + ... + 3^2 + 3
= [3^(k+2) - 3] / 2
T(n = 3^k) = F(k) = (9 * n - 3) / 2 = O(n) T(n = 3^k) = F(k) = (9 * n - 3) / 2 = O(n)
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