LeetCode - Pow(x, n)

public double myPow(double x, int n) {
    if(x == 1)
        return x;
    if(x == -1)
        return n % 2 == 0 ? 1: -1;
    if(n == Integer.MAX_VALUE || n == Integer.MIN_VALUE)
        return 0;
    if(n == 0)
        return 1;
    if(n<0){
        n = -n;
        x = 1/x;
    }
    return (n%2 == 0) ? myPow(x*x, n/2) : x*myPow(x*x, n/2);
}