剑指offer-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。

全面高效的递归算法：a^n=(a^n/2)*(a^n/2)

public class Solution {    public double Power(double base, int exponent) {        if(0==exponent)        return 1;     if(exponent > 0){if (1 == exponent)return base;double result = Power(base, exponent >> 1);result *= result;if (1 == (exponent & 0x1))result *= base;return result;    }else {             double nums = base;            int flag = -exponent;            for(int i = 1; i < flag; i++){                nums = nums * base;            }            return 1/nums;            }  }}