剑指offer--数值的整数次方

public class Solution {
    public double Power(double base, int exponent) {
       double result = 1.0;
        if ( exponent == 0.0){
           return 1.0;
       }else if ( exponent > 0){
           for(int i = 0 ; i < (int)(exponent) ; i++){
               result *= base;
           }
       }else{
           if ( base == 0){
               throw new RuntimeException("分母不能为0");
           }
           exponent = -exponent;
           for ( int i = 0 ; i < exponent ; i++){
               result *= base;
           }
           result = 1/result;
       }
        return result;
    }
}

#include <cmath>
using namespace std;

class Solution {
    public:
        int equal(double a,double b){
            return fabs(a-b) < 0.000001;
        }

        double Power(double base, int exponent) {
            # 0的0次幂没有意义,因此要把base为0的情况排除在外
            # 浮点数的相等需要利用函数，而不是base == 0.0这样的语句 
            if(exponent == 0 && !this->equal(base,0.0)){
                return 1.0;
            }
            int n = (exponent>0)?exponent:-exponent;
            double result = 1.0;
            for(int i = 1 ; i <= n ; i++){
                result *= base;
            }
            return (exponent>0)?result:1.0/result;
        }
};