《剑指offer》牛客网java题解-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。
double类型不能直接用==判断，底数等于0指数小于0时不能将0作为分母。
使用了优化的求幂函数

public class Solution {
    public double Power(double base, int exponent) {
        if (equal(base, 0) && exponent < 0) {
            return 0.0;
        }
        int absExponent = Math.abs(exponent);
        double result = powerWithExponent(base, absExponent);
        if (exponent < 0) {
            result = 1.0 / result;
        }
        return result;
    }

    public double powerWithExponent(double base, int absExponent) {
        if ( absExponent == 0 )
            return 1;

        if ( absExponent == 1 )
            return base;

        double result = powerWithExponent(base, absExponent >>1);
        result *= result;
        if ( (absExponent&1) == 1 )
            result *= base;

        return result;
    }

    public boolean equal(double num1, double num2) {
        if ((num1 - num2 > -0.0000001) && (num1 - num2 < 0.0000001)) {
            return true;
        } else
            return false;
    }
}