剑指offer-数值的整数次方

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。保证base和exponent不同时为0

class Solution {
public:
    
    double Power(double base, int exponent) {
        if(base==1.0) return 1.0;
        if(exponent==0) return 1.0;
        double temp=1.0;
        if(exponent>0)
        {
            while(exponent)
            {
                temp = temp*base;
                exponent--;
            } 
        }
        else{
            
            while(exponent)
            {
                temp =temp/base;
                exponent++;
            } 
        }
        return temp;
    }
    
};
原文地址:https://www.cnblogs.com/loyolh/p/12347092.html