以下哪一项正确确定了 T(1) = 1 的递推关系的解?
T(n) = 2T(n/2) + Logn (A) Θ(n)
(B) Θ(nLogn)
(C) Θ(n*n)
(D) Θ(log n)答案：(一)
解释：
T(n) = 2T(n/2) + log n
T(1) = 1
Substitute n = 2^k
T(2^k) = k + 2T(2^(k-1))
T(2^k) = k + 2(k-1) + 4T(2^(k-2))
= k + 2(k-1) + 4(K-2) + 8T(2^(k-3))
= k + 2(k-1) + 4(K-2) + 8(k-3) + 16T(2^(k-4))
= k + 2(k-1) + 4(K-2) + 8(k-3) + ...... + 2^kT(2^(k-k))
= k + 2(k-1) + 4(K-2) + 8(k-3) + .......+ 2^kT(1)
= k + 2(k-1) + 4(K-2) + 8(k-3) + .......+ 2^k --------(1)
2T(2^k) = 2k + 4(k-1) + 8(K-2) + ...... + 2*2^k + 2^(k+1) --------(2)
Subtracting 1 from 2, we get below
T(2^k) = - k + 2 + 4 ...... 2^(k-2) + 2^(k-1) + 2^k + 2^(k+1)
= - k + 2 * (1 + 2 + 4 + ..... 2^k)
= -k + [2*(2^k - 1)] / [2-1]
= -k + [2*(2^k - 1)]
T(n) = -Logn + 2*(n - 1)
T(n) = Θ(n) 这个问题的测验