2019-11-02 12:07:41 +08:00
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# 16. 数值的整数次方
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[NowCoder](https://www.nowcoder.com/practice/1a834e5e3e1a4b7ba251417554e07c00?tpId=13&tqId=11165&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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给定一个 double 类型的浮点数 base 和 int 类型的整数 exponent，求 base 的 exponent 次方。
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## 解题思路
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下面的讨论中 x 代表 base，n 代表 exponent。
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<!--<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>-->
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2019-12-06 00:29:51 +08:00
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<div align="center"> <img src="pics/48b1d459-8832-4e92-938a-728aae730739.jpg" width="330px"> </div><br>
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2019-11-02 12:07:41 +08:00
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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)。
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```java
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public double Power(double base, int exponent) {
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if (exponent == 0)
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return 1;
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if (exponent == 1)
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return base;
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boolean isNegative = false;
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if (exponent < 0) {
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exponent = -exponent;
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isNegative = true;
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}
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double pow = Power(base * base, exponent / 2);
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if (exponent % 2 != 0)
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pow = pow * base;
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return isNegative ? 1 / pow : pow;
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}
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```
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2019-11-02 14:39:13 +08:00
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2019-11-02 17:33:10 +08:00
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
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