剑指offer：数值的整数次方 代码实现

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。

#include <stdio.h>#include <iostream>#include <vector>using namespace std;class Solution {public:    double Power(double base, int exponent) {        int index = 0;        double mul = 1.0;               /* 如果exponent = 0 输出1 */        if(exponent == 0)        {            return 1.00000;        }        /* 如果base = 0 输出1 */        if(base >= -0.000001 && base <= 0.000001)        {            return 0;        }        /* 如果指数大于0 */        if(exponent > 0)        {            for(index = 0; index < exponent; index++)            {                mul *= base;            }        }        else        {            exponent = -exponent;            for(index = 0; index < exponent; index++)            {                mul *= base;            }            mul = 1.0/mul;        }        return mul;    }};int main(){    Solution s;    cout << s.Power(0,0) << endl;    return 0;}