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题目描述：

       实现函数double Power(double base, int exponent),求base的exponse次方。不得使用库函数，同时不需要考虑大数问题。

解题思路：

       首先从数学的定义去分析该题：

①任何非0的0次方等于1.

②base^exponent当exponent为负数值有base^exponent=1.0/(base^(-exponent)) 此时base为0，exponent为负数为无效输入。

需要注意:

①判断base等于0时不能直接判断，float、double用计算机表示是有误差的。

②三种错误处理的方法

解法一：

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

/*
 * @name   Equal
 * @brief  判断两个浮点数是否相等
 * @param  [in] double lfValue1
 * @param  [in] double lfValue2
 * @return bool
 */
 bool Equal(double lfValue1, double lfValue2)
{
    if (lfValue1 - lfValue2 > -0.0000001
        && lfValue1 - lfValue2 < 0.0000001)
    {
        return true;
    }
    return false;
}

 /*
  * @name   PowerUnsignedExponent
  * @brief  求base^exponent (exponent为正数)
  * @param  [in] double base
  * @param  [in] int exponent
  * @return double
  */
  double PowerUnsignedExponent(double base, int exponent)
 {
     double lfResult = 1.0;
     while (exponent--)
     {
         lfResult *= base;
     }
     return lfResult;
 }

  bool g_bInvalidInput = false;

/*
 * @name   Power
 * @brief  求base的exponent次方
 * @param  [in] double base
 * @param  [in] int exponent
 * @return double
 */
 double Power(double base, int exponent)
 {
     g_bInvalidInput = false;
     if (Equal(base, 0.0) && exponent < 0)
     {
         g_bInvalidInput = true;
         return 0.0;
     }

     if (Equal(base, 0.0))
     {
         return 0.0;
     }

     if (exponent == 0)
     {
         return 1.0;
     }

    double lfResult = 1.0;
    bool bPositive = exponent >= 0 ? true : false;
    if (!bPositive)
    {
        exponent = -exponent;
    }
    lfResult = PowerUnsignedExponent(base, exponent);
    if (!bPositive)
    {
        lfResult = 1.0 / lfResult;
    }
    return lfResult;
}

int _tmain(int argc, _TCHAR* argv[])
{
    //测试正常值 -2 和 2 的平方和立方
    cout << "2^2=" << Power(2.0, 2) << endl;
    cout <<"error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;
    cout << "-2^2=" << Power(-2.0, 2) << endl;
    cout << "error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;
    cout << "2^3=" << Power(2.0, 3) << endl;
    cout << "error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;
    cout << "-2^3=" << Power(-2.0, 3) << endl;
    cout << "error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;

    //测试exponent为负值
    cout << "2^-3=" << Power(2.0, -3) << endl;
    cout << "error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;
    cout << "-2^3=" << Power(-2.0, -3) << endl;
    cout << "error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;

    //测试exponent为负值
    cout << "0^-3=" << Power(0, -3) << endl;
    cout << "error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;
    cout << "-3^0=" << Power(-3.0, 0) << endl;
    cout << "error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;

    //大数测试
    cout << "2^32=" << Power(2.0, 32) << endl;
    cout << "error:" << (g_bInvalidInput == false ? "no error" : "invalid") << endl;

    system("pause");
    return 0;
}

高效解法：

相关知识：

        位运算效率高于（*）乘、（\）除、（%）取余

只需改变上面解法的PowerUnsignedExponent函数：

 /*
  * @name   PowerUnsignedExponent
  * @brief  求base^exponent (exponent为正数)
  * @param  [in] double base
  * @param  [in] int exponent
  * @return double
  */
  double PowerUnsignedExponent(double base, int exponent)
 {
     if (exponent == 1)
     {
         return base;
     }   
     //右移一位代替除2操作
     double lfResult = PowerUnsignedExponent(base, exponent >> 1);
     lfResult *= lfResult;
     //使用位与操作判断奇、偶
     if (exponent & 1 == 1)
     {
         lfResult *= base;
     }
     return lfResult;
 }

运行结果：