幂运算 power()

class Solution {
public:
    double myPow(double x, int n) {
        bool down = true;        // 负数次方时为倒数
        bool sign = true;        // 
        
        if (n < 0)
            down = false;
        
        if (n > INT_MAX || n < INT_MIN )
            return 0;
        
        if (n == INT_MIN)
        {
            sign = false;
            n = INT_MAX;
        }
        
        if (x == 0 || x == 1 || n == 1)
            return x;
        
        int m = abs(n);
        if (x == -1)
        {
            if ( ((m%2 == 0) && sign) || ((m%2 == 1) && !sign) )
                return 1;
            return -1;
        }

        if (m == 0)
        {
            return 1;
        }
        else
        {
            while ((m & 1) == 0)           // 过滤掉二进制中末尾的0
            {
                m >>= 1;
                x *= x;
            }
        }  
       
        double ret = 1;
        while (m != 0)
        {
            if ((m & 1) != 0)
                ret *= x;
            x *= x;
            m >>= 1;
        }

        if (!down)
            ret = 1 / ret;
        if (!sign)
            ret *= ret;
        return ret;
    }
};