数值的整数次方

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。

 

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