【leetcode刷题笔记】Pow(x, n)

 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 }