LeetCode算法题之Pow(x, n)

问题描述：

Implement pow(x, n).自己写函数实现乘方运算

解题思路：

直接实现肯定超时，并且要考虑指数为0，指数为1，指数为负，底数为0的特殊情况，此题用递归方法实现，

以下代码来自《剑指offer》，此代码写的太赞了，如果想详细了解解题思路请买书！！刚看完所以直接贴上去运行了，必然轻松A过，向大神致敬！！！

bool g_InvalidInput = false;bool equal(double num1, double num2);double PowerWithUnsignedExponent(double base, unsigned int exponent);class Solution{public:    double pow(double base, int exponent)    {        g_InvalidInput = false;        if(equal(base, 0.0) && exponent < 0)//默认一定范围外的数约等于0        {            g_InvalidInput = true;            return 0.0;        }        unsigned int absExponent = (unsigned int)(exponent);        if(exponent < 0)//指数小于0            absExponent = (unsigned int)(-exponent);        double result = PowerWithUnsignedExponent(base, absExponent);//递归实现        if(exponent < 0)            result = 1.0 / result;        return result;    }    double PowerWithUnsignedExponent(double base, unsigned int exponent)    {        if(exponent == 0)            return 1;        if(exponent == 1)            return base;        double result = PowerWithUnsignedExponent(base, exponent >> 1);//此处exponent做的是移位运算，比直接除2快的多，大神！！！        result *= result;        if((exponent & 0x1) == 1)            result *= base;        return result;    }    bool equal(double num1, double num2)    {        if((num1 - num2 > -0.0000001)                && (num1 - num2 < 0.0000001))            return true;        else            return false;    }};