判断以下T(n) = 2T(n/2) + nlogn的时间复杂度是否为Θ(𝑛𝑙𝑜𝑔2𝑛)

不是。

根据主定理，T(n) = 2T(n/2) 的情况属于第二种情况，即a=2, b=2, f(n)=n^1，因为logb(a) = log2(2) = 1，所以f(n) = n^1 = n^logb(a)，因此时间复杂度为Θ(nlogn)。并不是Θ(𝑛𝑙𝑜𝑔2𝑛)。

相关问题

Can T(n) = 2T(n/2) + nlogn use master theorem?

Yes, T(n) = 2T(n/2) + nlogn can be solved using the master theorem.

First, we identify the values of a, b, and f(n) in the recurrence relation:

a = 2 (the number of recursive calls)
b = 2 (the size of the subproblems)
f(n) = nlogn (the cost of dividing the problem into subproblems and combining their solutions)

Next, we compute the value of logb(a):

log2(2) = 1

Since f(n) = nlogn is asymptotically larger than n^(log2(2)) = n, we can apply case 2 of the master theorem:

If f(n) is Θ(n^c) for some constant c and a >= b^c, then T(n) is Θ(n^c logn).

In this case, c = 1 because f(n) = nlogn.

Since a = 2 and b^c = 2^1 = 2, we have a > b^c and can apply case 2:

T(n) is Θ(nlogn).

Therefore, the solution to the recurrence relation T(n) = 2T(n/2) + nlogn is T(n) = Θ(nlogn).

T(n) = 2T(n/2) + nlogn的asymptotically tight bound

我们可用主定理（Master Theorem）来证明T(n) = 2T(n/2) 的时间复杂度为nlogn的asymptotically tight bound。

根据主定理，对于递归式T(n) = aT(n/b) + f(n)（其中a和b为常数，且f(n)为渐进正的函数）：

如果f(n) = O(n^log[b]a-ε) 对某个常数ε > 0，则T(n) = Θ(n^log[b]a)。
如果f(n) = Θ(n^log[b]a * logk n)，则T(n) = Θ(n^log[b]a * logk+1 n) 。
如果f(n) = Ω(n^log[b]a+ε) 对某个常数ε > 0，并且如果对某个常数c < 1和足够大的n，有af(n/b) ≤ cf(n)，则T(n) = Θ(f(n)) 。

在这里，a = 2，b = 2，f(n) = 0，因此f(n) = Θ(n^0)，即 f(n) = Θ(1)。

因为 f(n) = Θ(1) = Θ(n^0)，所以我们可以应用主定理的第一种情况来计算时间复杂度，即：

T(n) = Θ(n^log[b]a) = Θ(n^log[2]2) = Θ(nlogn)

证毕。