【leetcode刷题笔记】Pow(x, n)

Implement pow(x, n).

1. 普通的递归把n降为n-1会超时，要用二分的方法，每次把xⁿ = x^[n/2] * x^[n/2] * x^n-[n/2]*2, [n/2]表示n除以2下取整。
2. n有可能取负数，负数的时候，先计算pow(x,-n)，然后返回1/pow(x,-n);

代码如下：

 1 public class Solution {
 2     public double pow(double x, int n) {
 3         if(n == 0)
 4             return 1;
 5         if(n == 1)
 6             return x;
 7
 8         boolean isNeg = false;
 9
10         //if n is negative,we first caculate 1/pow(x,n) = pow(x,-n)
11         if(n<0){
12             isNeg = true;
13             n *= -1;
14         }
15
16         //binary
17         int mid = n /2;
18         int left = n - mid*2;
19         double front = pow(x, mid);
20         double end = pow(x, left);
21
22         if(isNeg)
23             return 1/(front*front*end);
24         else {
25             return front*front*end;
26         }
27     }
28 }

【leetcode刷题笔记】Pow(x, n)