LeetCode 50:Pow(x, n)

class Solution {
    public double myPow(double x, int n) {
        if (x == 0.0) {
            return x;
        }
        
        double sign = n > 0 ? 1.0 : -1.0;
        long nn = Math.abs(Long.valueOf(n));
        
        double result = 1.0;
        
        while (nn > 0) {
            if (nn % 2 == 1) {
                result *= x;
            }
            x *= x;
            nn /= 2;
        }
        return sign > 0 ? result : 1.0 / result;
    }
}