11 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。

比如求2的32次方，先求出2的16次方，然后2^16 * 2^16
然后递归求2^16，如果次方为奇数，则再乘一个base

C++：

class Solution {
private:
    bool invalidInput = false ;
public:
    bool Equal(double num1 , double num2){
        if ((num1-num2)<0.0000001 && (num1-num2)>-0.0000001)
            return true ;
        else
            return false ;
    }

    double PowerWithUnsignedExponent(double base , int exponent){
        if (exponent == 0)
            return 1 ;
        if (exponent == 1)
            return base ;
        double result = PowerWithUnsignedExponent(base,exponent>>1) ;
        result *=result ;
        if (exponent&1)
            result *= base ;
        return result ;
    }

    double Power(double base , int exponent){
        bool invalidInput = false ;
        if (Equal(base,0.0) && exponent < 0){
            invalidInput = true ;
            return 0.0 ;
        }
        int absExponent = abs(exponent) ;
        double result = PowerWithUnsignedExponent(base,absExponent) ;
        if (exponent < 0)
            result = 1.0/result ;
        return result ;
    }
};

java:

public class Solution {
    public double Power(double base, int exponent) {
        if (exponent == 0)
            return 1 ;
        if (exponent == 1)
            return base ;
        boolean flag = false ;
        if (exponent < 0){
            exponent = -exponent ;
            flag = true ;
        }
        double result = Power(base * base , exponent/2) ;
        if (exponent%2 != 0){
            result *= base ;
        }
        if (flag){
            result = 1/result ;
        }
        return result ;
  }
}