求解递归时间复杂度

迭代法（递归树）

每一次对过程的重复称为一次迭代，而每一次迭代得到的结果会作为下一次迭代的初始值。重复执行一系列运算步骤，从前面的量依次求出后面的量的过程。

例1

problem

$$T(n)=2\times T(\frac{n}{4})+\sqrt{n},T(1)=1$$

solution

$$T(n)=2\times T(\frac{n}{4})+\sqrt{n}$$

$$T(n)=2\times (2\times T(\frac{n}{16})+\sqrt{\frac{n}{4}})+\sqrt{n}$$

$$T(n)=4\times T(\frac{n}{16})+2\times \frac{\sqrt{n}}{2}+\sqrt{n}$$

$$T(n)=4\times T(\frac{n}{16})+2\sqrt{n}$$

$$\dots$$

$$T(n)=2^k\times T(\frac{n}{2^{2k}})+k\sqrt{n}$$

令 $2^{2k}=n$，即 $2^k=\sqrt{n}$，$k={\log }_2^{\sqrt{n}}={\log }_2^{n^{\frac{1}{2}}}=\frac{1}{2}{\log }_2^n$

$$T(n)=\sqrt{n}\times T(1)+\frac{1}{2}{\log }_2^n\sqrt{n}$$

$$T(n)=\sqrt{n}+\frac{1}{2}\sqrt{n}{\log }_2^n$$

所以时间复杂度为 $O(\sqrt{n}\log n)$

例2

problem

$$T(n)=4\times T(\frac{n}{2})+n^2{\log }^{2}n,T(1)=1$$

solution

$$T(n)=4\times (4\times T(\frac{n}{4})+(\frac{n}{2}{)}^2{\log }^2(\frac{n}{2}))+n^2{\log }^{2}n$$

$$T(n)=16\times T(\frac{n}{4})+n^2{\log }^2(\frac{n}{2})+n^2{\log }^{2}n$$

$$T(n)=16\times T(\frac{n}{4})+n^2\times ({\log }^2(\frac{n}{2})+{\log }^{2}n)$$

$$T(n)=16\times T(\frac{n}{4})+n^2\times ({\log }^{2}n+{\log }^{2}n-{\log }^{2}2)$$

$$\dots$$

$$T(n)=2^{2k}\times T(\frac{n}{2^k})+n^2\times (k{\log }^{2}n-(1+4+...+{\log }^2{2}^{k-1}))$$

$$T(n)=2^{2k}\times T(\frac{n}{2^k})+n^2\times (k{\log }^{2}n-\sum _{i=1}^{k-1}{i}^2)$$

$$T(n)=2^{2k}\times T(\frac{n}{2^k})+n^2\times (k{\log }^{2}n-\frac{(k-1)(k-2)(2\times (k-1)-1)}{6})$$

$$T(n)=2^{2k}\times T(\frac{n}{2^k})+n^2\times (k{\log }^{2}n-\frac{2k^3-9k^2+13k-6}{6})$$

令 $2^k=n$ ，即 $2^{2k}=n^2$ ， $k=logn$

$$T(n)=n^2\times T(1)+n^{2}lo{g}^{3}n-n^2\times \frac{2{\log }^{3}n-9{\log }^{2}n+13logn-6}{6}$$

所以时间复杂度为 $O(n^{2}lo{g}^{3}n)$

例3

problem

$$T(n)=3\times T(\frac{n}{4})+n\log n$$

solution

$$T(n)=3\times (3\times T(\frac{n}{16})+\frac{n}{4}\log (\frac{n}{4}))+n\log n$$

$$T(n)=9\times T(\frac{n}{16})+\frac{3n}{4}\log (\frac{n}{4})+n\log n$$

$$T(n)=27\times T(\frac{n}{64})+\frac{9n}{16}\log (\frac{n}{16})+\frac{3n}{4}\log (\frac{n}{4})+n\log n$$

$$T(n)=27\times T(\frac{n}{64})+n\times (\log n\times (\frac{9}{16}+\frac{3}{4}+1)-(\frac{9}{16}\log 2^4+\frac{3}{4}\log 2^2))$$

$$T(n)=27\times T(\frac{n}{64})+n\times (\log n\times (\frac{9}{16}+\frac{3}{4}+1)-(\frac{9}{4}\log 2+\frac{3}{2}\log 2))$$

$$T(n)=27\times T(\frac{n}{64})+n\times (\log n\times (\frac{9}{16}+\frac{3}{4}+1)-(\frac{9}{4}+\frac{3}{2}))$$

$$\dots$$

$$T(n)=3^k\times T(\frac{n}{2^{2k}})+n\times (\log n\times (-4\times ((\frac{3}{4}{)}^k-1)))-3\times ((\frac{3}{2}{)}^{k-1}-1))$$

令 $2^{2k}=n$，即 $k={\log }_{4}n$

$$T(n)=3^{{\log }_{4}n}+n\times (\log n\times (-4\times (\frac{3}{4}{)}^{{\log }_{4}n}+4)-3\times (\frac{3}{2}{)}^{{\log }_{4}n-1}+3)$$

$$T(n)=n^{{\log }_{4}3}+n\log n\times (-4\times (\frac{3}{4}{)}^{{\log }_{4}n}+4)-3n\times (\frac{3}{2}{)}^{{\log }_{4}n-1}+3n$$

$$T(n)=n^{{\log }_{4}3}+n\log n\times (-4\times n^{{\log }_4\frac{3}{4}}+4)-2n\times (\frac{3}{2}{)}^{{\log }_{4}n}+3n$$

$$T(n)=n^{{\log }_{4}3}+n\log n\times (-4\times n^{{\log }_4\frac{3}{4}}+4)-2n\times n^{{\log }_4\frac{3}{2}}+3n$$

$$T(n)=n^{{\log }_{4}3}+4n\log n-4n\log n\times n^{{\log }_4\frac{3}{4}}-2n\times n^{{\log }_4\frac{3}{2}}+3n$$

$$T(n)=n^{{\log }_{4}3}+4n\log n-4n\log n\times n^{{\log }_{4}3-{\log }_{4}4}-2n\times n^{{\log }_{4}3-{\log }_{4}2}+3n$$

$$T(n)=n^{{\log }_{4}3}+4n\log n-4n^{{\log }_{4}3}\log n-\frac{2n\times n^{{\log }_{4}3}}{\sqrt{n}}+3n$$

$$T(n)=n^{{\log }_{4}3}+4n\log n-4n^{{\log }_{4}3}\log n-2\sqrt{n}\times n^{{\log }_{4}3}+3n$$

所以时间复杂度为 $O(n\log n)$