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;
}
}