《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;    }}