Leetcode: Pow(x, n)

class Solution 
{
private:
    double power(double x, int n)    
    {
        if (0 == n) return 1;

        double v = power(x, n / 2);
        // 如果n是偶数，则返回x^(n/2)*x^(n/2)
        if (0 == n % 2) return v * v;
        // 如果n是奇数，则返回x^(n/2)*x^(n/2)*x^(n%2)
        else return v * v * x;
    }
public:
    double myPow(double x, int n) 
    {
        if (n < 0) return 1.0 / power(x, -n);
        else return power(x, n);
    }
};