数值的整数次方

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题目描述

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。思路：要考虑到指数为负数的情形！如果指数是负数，就先求其绝对值，求出n次方后再求倒数

public class Solution {
    double sum = 0.0;
    public double Power(double base, int exponent) {
        if(equal(base,0.0)&&exponent < 0){//如果底数等于0 指数小于0 则返回0.0
            sum = 0.0;
        }//if
        else{
        if(exponent>=0){//如果exponent>=则直接调用fun()函数
           sum = fun(base,exponent);
        }//if
            else{
            exponent = Math.abs(exponent);
            sum = (double)(1/fun(base,exponent));
        }//else
     }//else
       return sum; 
  }//Power
    private double fun(double base,int exponent){
        double s = 1.0;
        for(int i = 0;i < exponent;i++){
            s = s*base;
        }
        return s;
    }//fun
    private boolean equal(double num1,double num2){
        if((num1 - num2 > -0.00000001)&&(num1 - num2 < 0.00000001))
            return true;
        else return false;
    }//equal
}