Pow(x, n)

Implement pow(x, n), which calculates x raised to the power n (xn).

Example 1:

Input: 2.00000, 10
Output: 1024.00000
Example 2:

Input: 2.10000, 3
Output: 9.26100

该题简单的循环累乘会超时，时间复杂度O(n).不够快！
可以用分治法将复杂度降低到O(logn)
需要注意数值溢出问题！

class Solution {
public:
    double myPow(double x, int n) {
        return helper(x,n);
    }
    
    double helper(double x, long long n)
    {
        if(n<0) return 1/helper(x,-n);
        if (n==0) return 1;
        if (n%2==0)
        {
            double tmp = helper(x,n/2);
            return tmp*tmp;
        }
        else 
        {
            double tmp =helper(x,(n-1)/2);
            return tmp*tmp*x;
        }
    }
    
};