剑指offer 12 ： 数值的整数次方

1 考虑指数和底数不同的条件

#include <iostream>
#include <cmath>

using namespace std;


class solution
{
public:
    //12 数值的整数次方
    double power(double base, int exponent) {
        if (exponent == 0) {
            return 1;//指数为0
        }
        if (Equal(base, 0.0) == true && exponent < 0) {
            cout << "error"  << endl;
            return 0.0;
        }//底数为0

        double res = 1.0;
        if (exponent > 0) {
            res = PowerNormal (base, exponent);//指数大于0
        }
        else {
            res = PowerNormal (base, -exponent);//指数小于0
            res = 1/res;
        }

        return res;
    }
private:
    double PowerNormal(double base, int exponent) {
        double res = 1;
        for (int i = 0; i < exponent; i++) {
            res = res * base;
        }
        return res;
    }//求次方

    bool Equal(double left ,double right) {
        if ((left - right) == 0.000000) {
            return true;
        }
        else {
            return false;
        }
    }//比较left与right是否相等
};

int main()
{
    solution solu;
    cout << solu.power(2, -3);

    return 0;
}