12 求数值的整数次方

//题目：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。
//分情况考虑：
//第一种情况：底数是0，且指数是负数；
//第二种情况：指数小于0；
class Solution 
{
public:
    bool g_InvalidInput = false;
    double Power(double base, int exponent) 
    {
        //第一种情况：
        if (Equal(base, 0.0) && exponent < 0)
        {
            g_InvalidInput = true;
            return 0.0;
        }
        //第二种情况：考虑指数问题,指数小于0，就要取绝对值
        unsigned int absExponent = (unsigned int ) exponent;
        if (exponent < 0)
        {
            absExponent = (unsigned int)(-exponent);
        }
        double result = PowerWithExponent(base, absExponent);
        if(exponent < 0)
        {
            result = 1.0 /result;
        }
        return result;
    }
private:
    bool Equal(double num1, double num2)
    {
        if(num1 - num2 > -0.000001 && num1-num2 <0.000001)
        {
            return true;
        }
        return false;
    }
    double PowerWithExponent(double base, unsigned int num)
    {
        if (num == 0)
        {
            return 1;
        }
        if (num == 1)
        {
            return base;
        }
        //***重点理解***由于要递归，故要定义返回值result，终止条件为num == 1
        double result = PowerWithExponent(base, num>>1);
        result *= result;
        if (num & 0x1 == 1)
        {
            result *= base;
        }
        return result;
    }
};