剑指offer 数值的整数次幂，求base的exponent次方，不能使用库函数，同时不需要考虑大数问题

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/** * Created by zj on 2017/3/13 0013.

数值的整数次幂，求base的exponent次方，不能使用库函数，同时不需要考虑大数问题 */public class Exponent {    public double power(double base,int exponent)throws Exception{        double result = 0.0;        if(equals(base,0.0) && exponent < 0){            throw new Exception("0的负次幂没有意义！");        }else if(exponent < 0){            return 1/powerWithExponent(base,-exponent);        }else{            return powerWithExponent(base,exponent);        }    }    public double powerWithExponent(double base,int exponent){        if(exponent == 0){            return 1;        }        if(exponent == 1){            return base;        }        double result = 1.0;        for(int i =0;i<exponent;i++){            result = result * base;        }        return result;    }    public boolean equals(double num1,double num2){        if((num1-num2 < 0.00000001)&&(num1-num2 >-0.00000001)){            return true;        }else{            return false;        }    }    public static void main(String[] args) {        Exponent ex = new Exponent();        try{            System.out.print(ex.power(3.0,-3));        }catch(Exception e){            System.out.print("Exception");        }    }}

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