Pow(x, n)

/**
 * 50. Pow(x, n)<br>
 * 快速幂<br>
 * https://leetcode-cn.com/problems/powx-n/<br>
 * 
 * @author wuwang
 *
 */
public class p50 {
	public double myPow(double x, int n) {
		if (n == 0)
			return 1;
		double r = 1;
		long absn = Math.abs((long) n);// 很多時候对于int范围一定要注意 -2147483648 int_min 求abs直接取负数就會溢出
		while (absn > 0) {
			if (absn % 2 == 1)
				r *= x;
			x *= x;
			absn >>= 1;
		}
		if (n < 0)
			return 1 / r;
		return r;
	}
}