LeetCode 50 Pow(x, n)

Implement pow(x, n).

/**
 * A constant holding the minimum value an {@code int} can
 * have, -2<sup>31</sup>.
 */
@Native public static final int   MIN_VALUE = 0x80000000;
/**
 * A constant holding the maximum value an {@code int} can
 * have, 2<sup>31</sup>-1.
 */
@Native public static final int   MAX_VALUE = 0x7fffffff;

int的表示范围是[-2³¹, 2³¹-1]，当n为MIN_VALUE时，取反会溢出。

	public double myPow(double x, int n) {
		if (n == Integer.MIN_VALUE) //-INT_MIN will cause overflow
			return myPow(x, n + 1) / x;
		if (n < 0) return 1 / myPow(x, -n);
		if (n == 0) return 1;
		double halfResult = myPow(x, n >> 1);
		return n % 2 == 0 ? halfResult * halfResult : halfResult * halfResult * x;
	}