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docs/notes/剑指 offer 题解.md
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@ -442,7 +442,9 @@ public int pop() throws Exception {
求斐波那契数列的第 n 项，n <= 39。
<div align="center"><img src="https://latex.codecogs.com/gif.latex?f(n)=\left\{\begin{array}{rcl}0&&{n=0}\\1&&{n=1}\\f(n-1)+f(n-2)&&{n>1}\end{array}\right."/></div> <br>
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?f(n)=\left\{\begin{array}{rcl}0&&{n=0}\\1&&{n=1}\\f(n-1)+f(n-2)&&{n>1}\end{array}\right."/></div> <br> -->
<div align="center"> <img src="pics/45be9587-6069-4ab7-b9ac-840db1a53744.jpg"/> </div><br>
## 解题思路
@ -896,7 +898,9 @@ public int NumberOf1(int n) {
下面的讨论中 x 代表 base，n 代表 exponent。
<div align="center"><img src="https://latex.codecogs.com/gif.latex?x^n=\left\{\begin{array}{rcl}(x*x)^{n/2}&&{n\%2=0}\\x*(x*x)^{n/2}&&{n\%2=1}\end{array}\right."/></div> <br>
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?x^n=\left\{\begin{array}{rcl}(x*x)^{n/2}&&{n\%2=0}\\x*(x*x)^{n/2}&&{n\%2=1}\end{array}\right."/></div> <br>-->
<div align="center"> <img src="pics/48b1d459-8832-4e92-938a-728aae730739.jpg"/> </div><br>
因为 (x\*x)<sup>n/2</sup> 可以通过递归求解，并且每次递归 n 都减小一半，因此整个算法的时间复杂度为 O(logN)。