12、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。

 

解题思路：首先要注意特殊情况的处理   base为0情况  exponent为0 为1情况  exponent为负数情况

#include <iostream>
using namespace std;
class Solution {
public:
    //要注意的几点
    //1、base == 0
    //2、exponent == 0；==1
    //3、exponent 为负数
    double Power(double base, int exponent) {
        if(equal(base,0.0))
        {
            return 0.0;
        }
        unsigned int abExponent = (unsigned int)(exponent);//一定要注意此处给其赋值，否则如果exponent不小于0时，该值随机，会出现错误
        if(exponent < 0)
            abExponent = (unsigned int)(-exponent);
        double result = 1.0;
        for(int i=1;i<=abExponent;i++)
        {
            result *= base;
        }
        if(exponent<0)
            return 1.0/result;
        else
            return result;


    }
    bool equal(double num1,double num2)
    {
        if((num1-num2)>-0.0000001 && (num1-num2)<0.0000001)
        {
            return true;
        }
        else
            return false;
    }
    //方法二：a^n = a^(n/2)*a^(n/2)   n为偶数
    //        a^n = a^(n/2)*a^(n/2)*a n为奇数
    double Power2(double base, int exponent) {
        if(equal(base,0.0))
        {
            return 0.0;
        }
        if(exponent == 0)
            return 1.0;
        else if(exponent == 1)
            return base;
        unsigned int abExponent = (unsigned int)(exponent);
        if(exponent < 0)
            abExponent = (unsigned int)(-exponent);
        double result = Power2(base,abExponent>>1);//右移1相当于除以2
        result *= result;//a^(n/2)*a^(n/2);
        if(abExponent & 0x1 == 1)//奇数
        {
            result *= base;
        }
        if(exponent < 0)
            return 1.0/result;
        else
            return result;
    }
};
int main()
{
    double n ;
    int x;
    while(cin>>n>>x)
    {
        Solution s;
        cout<<"调用方法一:"<<n<<"的"<<x<<"次方为："<<s.Power(n,x)<<endl;
        cout<<"调用方法二:"<<n<<"的"<<x<<"次方为："<<s.Power2(n,x)<<endl;
    }
    return 0;
}

程序运行结果：