50. Pow(x, n)

package Leetcode.MathGeometry;

/**
 * Author: Xingxing Zheng
 * Date: 2024/12/16
 */
public class PowXN {

    public static double myPow(double x, int n) {
        // Handle edge cases where the base is 0 or the exponent is 0
        if (x == 0) {
            return 0; // Any number raised to a power with base 0 is 0
        }
        if (n == 0) {
            return 1; // Any number raised to the power of 0 is 1
        }

        // Convert n to long to handle Integer.MIN_VALUE edge case
        long N = n;
        if (N < 0) {
            // If the exponent is negative, take the reciprocal of the base
            x = 1 / x;
            N = -N; // Make the exponent positive
        }

        // Initialize the result to 1
        double result = 1.0;

        // Implement fast exponentiation
        while (N > 0) {
            // If the current bit of N (binary representation) is 1
            if (N % 2 == 1) {
                result *= x; // Multiply the current base value (x) into the result
            }

            // Update the base value by squaring it
            // This allows us to compute powers of 2 efficiently (e.g., x^2, x^4, x^8, etc.)
            x *= x;

            // Shift N to the right by 1 bit (equivalent to dividing by 2)
            // This moves to the next higher power of 2 in the binary representation of N
            N /= 2;
        }

        return result; // Return the final result
    }

    public static void main(String[] args) {
        // Test cases
        System.out.println("2^10 = " + myPow(2, 10));   // Output: 1024.0
        System.out.println("2^-2 = " + myPow(2, -2));   // Output: 0.25
        System.out.println("0^5 = " + myPow(0, 5));     // Output: 0.0
        System.out.println("5^0 = " + myPow(5, 0));     // Output: 1.0
        System.out.println("(-2)^3 = " + myPow(-2, 3)); // Output: -8.0
        System.out.println("(-2)^-3 = " + myPow(-2, -3)); // Output: -0.125
    }
}