leetcode-50. pow(x,n)

public double myPow(double x, int n)
{
    if(n == 0) return 1D;
    long N = n; //use long to avoid overflow.
    return solve(n < 0 ? (1 / x) : x, N < 0 ? (N * -1) : N);
}

public double solve(double x, long n)
{
    if(n == 1) return x;
    double val = solve(x, n / 2);
    return val * val * ((n % 2) == 0 ? 1 : x);
}