[LeetCode] Problem 50 - Pow(x, n)

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Implement pow(x, n), which calculates x raised to the power n (x^n).

Example

No.1

Input: 2.00000, 10

Output: 1024.00000

No.2

Input: 2.10000, 3

Output: 9.26100

No.3

Input: 2.00000, -2

Output: 0.25000

Explanation: 2^-2 = 1/2^2 = 1/4 = 0.25

Note

  • -100.0 < x < 100.0
  • n is a 32-bit signed integer, within the range [−231, 231 − 1]

Code

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public double myPow(double x, int n) {
    return n >= 0 ? pow(x, n) : 1 / pow(x, -n);
}

private double pow(double x, int n) {
    if (n == 0)
        return 1;

    double result = pow(x, n / 2);

    return n % 2 == 0 ? result * result : result * result * x;
}