剑指offer12

剑指offer12：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

解法一：分类讨论

class Solution {
public:
    double Power(double base, int exponent) {
        double answer = 1.0;
        if (exponent == 0)
            return 1.0;
        else if (base == 0.0)
            return 0.0;
        else if (exponent>0)
        {
            for (int i = 1; i <= exponent; i++)
            {
                answer *= base;
            }
        }
        else
        {
            base = 1 / base;
            for (int i = 1; i <= (0 - exponent); i++)
            {
                answer *= base;
            }
        }
        return answer;
    }
};

解法二：调用pow库函数

class Solution2 {
public:
    double Power(double base, int exponent) {
        return pow(base, exponent);
    }
};

main函数

#include "stdafx.h"
#include <iostream>
#include <vector>
using namespace std;

int main()
{
    double base;
    cout << "请输入浮点数base: " << endl;
    cin >> base;
    int exponent;
    cout << "请输入次幂：" << endl;
    cin >> exponent;
    Solution2 answer;
    cout << "答案是：" << answer.Power(base, exponent)<<endl;
    system("pause");
    return 0;
}