X的N次冪

class Solution {
    public double myPow(double x, int n) {
        double res = 1.0;
		for (int i = n; i != 0; i /= 2) {
			if (i % 2 != 0) {
				res *= x;
			}
			x *= x;
		}
		return n > 0 ? res : 1 / res;
    }
}