12数值的整数次方

题目描述

 
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。
 
思路:
使用递归,时间复杂度O(logn)
  • 当n为偶数,a^n =(a^n/2)*(a^n/2)
  • 当n为奇数,a^n = a^[(n-1)/2] * a^[(n-1)/2] * a
1 public class Solution {
2     public double Power(double base, int exponent) {
3         if(exponent==1) return base; 
4         if(exponent==-1) return 1/base; 
5         if(exponent==0) return 1; 
6         double res = Power(base,exponent>>1);
7         return (exponent &0x01)==0? res*res:base*res*res;
8     }
9 }

 

c++:20180807
 
 1 class Solution {
 2 public:
 3     double Power(double base, int exponent) {
 4             if(exponent==1) return base;
 5             if(exponent==-1) return 1/base;
 6             if(exponent==0) return 1;
 7             double res = Power(base,exponent>>1);
 8             if(exponent &1) return res*res*base;
 9             else return res*res;
10     }
11 };

 


 

posted @ 2017-11-06 09:27  乐乐章  阅读(179)  评论(0编辑  收藏  举报