剑指offer--面试题11: 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。

python实现：

# -*- coding:utf-8 -*-class Solution:    def Power(self, base, exponent):        if exponent==0:            return 1        if exponent<0:            return 1.0/self.Power(base, -exponent)        v = self.Power(base, exponent/2)        if exponent&1==0:#偶数            return v*v        else:            return base*v*v

c++实现：

class Solution {public:    double Power(double base, int exponent) {        if(exponent==0)            return 1;        if(exponent<0)            return 1/Power(base, -exponent);        double v = Power(base, exponent/2);        if(exponent&1==0)            return v*v;        else            return base*v*v;    }};