LeetCode 50: Pow(x, n)

Implement pow(x, n).

public class Solution {
    public double pow(double x, int n) {
        // Recursive problem
        // If n is odd number: pow(x,n) = x*pow(x,n/2)*pow(x,n/2);
        // If m is even number: pow(x,n) = pow(x,n/2)*pow(x,n/2);
        // Finally, if n is negative, we return the reciprocal of pow(x,-n)
        if (n < 0)
            return 1.0/power(x, -n);
        else
            return power(x, n);
    }
    
    private double power(double x, int n)
    {
        // Note: Don't miss this; Otherwise, entering into dead loop.
        if (n == 0)
            return 1.0;
            
        double v = power(x, n/2);
        
        if (n%2 == 0)
            return v*v;
        else
            return x*v*v;
    }
}