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2019-11-02 14:39:13 +08:00
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docs/notes/16. 数值的整数次方.md
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@ -12,7 +12,7 @@
<!--<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." class="mathjax-pic"/></div> <br>-->
<img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/48b1d459-8832-4e92-938a-728aae730739.jpg" width="330px">
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/48b1d459-8832-4e92-938a-728aae730739.jpg" width="330px"> </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)。
@ -34,3 +34,10 @@ public double Power(double base, int exponent) {
return isNegative ? 1 / pow : pow;
}
```
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-1.png"></img></div>