递归算法时间复杂度

递归算法时间复杂度分析

等比数列和公式:
$Sn = \frac{a_1(1-q^n)}{1-q}$

$Sn = \frac {a_1-a_nq}{1-q}$

$Sn = \frac{qa_n - a_1}{q-1}$

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example 1
$T(n)=2*T(\frac{n}{2})+n$

$T(n)=2*T(\frac{n}{2})+n\\ =n+n+n......+n\\=n*log_2n\\=O(n*log_2n)$

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example 2
$T(n) =2*T(\frac{n}{2}) +n^2$

$T(n) =2*T(\frac{n}{2}) +n^2\\=n^2+\frac{n^2}{2} +\frac{n^2}{4}+\frac{n^2}{8}......\frac{n^2}{2^{log_2n-1}}\\=n^2*\frac{1-(\frac{1}{2})^{log_2n}}{1-\frac{1}{2}}\\=2n^2(1-\frac{1}{n})\\=2n^2-2n\\=O(n^2)$

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Master method
$T(n)=a*T(\frac{n}{b})+f(n)$

case 1

Leaves grow faster than f
$f(n) = O(n^{log_b^{a-\delta}}) \\T(n)=\theta(n^{log_b^a})$
case 2

Leaves grow at the same rate as f
$f(n)=\theta(n^{log_b^a})\\T(n)=\theta(n^{log_b^a}*log_bn)$
case 3

f grows faster than leaves
$f(n) =\Omega(n^{log_b^{a+k}})\\T(n)=\theta(f(n))$

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Exercise
$T(n) = 4*T(\frac{n}{2})+n$

​ a = 4 b=2

​ $n^{log_b^a}=n^2$

​ 符合 case 1

​ $T(n)=\theta(n^2)$