Pow(x, n) @leetcode

Pow.java

package leetcode.binarySearch;

/**
 * Solution: BinarySearch  
 * 
 * Tricky:
 * 1. n could be negative.
 * 2. n cound be Integer.MIN_VALUE(-2147483648) and Integer.MAX_VALUE(2147483647) 
 * 3. handle n == 0, 1 as terminating condition.
 * 
 * @author jeffwan
 * @date Mar 10, 2014
 */
public class Pow {
	public double pow (double x, int n) {
		if (n == 0) {
			return 1;
		}
		
		if (n == 1) {
			return x;
		}
		
		if (n < 0) {
			// Value of pow(x, Integer.MIN_VALUE)? == pow(x, Integer.MAX_VALUE) * x
			if ( n == Integer.MIN_VALUE) {
				return 1 / (pow(x, Integer.MAX_VALUE) * x);
			} else {
				return 1 / pow(x, -n);
			}
		}
		
		// BinarySearch Thought
		double half = pow(x, n / 2); 
		if (n % 2 == 1) {
			return half * half * x;
		} else {
			return half * half;
		}
	}
	
	// Brute Force
	public double pow2 (double x, int n) {
		double temp = x;
		for (int i = 1 ; i<= n; i++) {
			x *= temp ;
		}
		return x;
	}
}