"Prove safety offer" - integer power values (Java)

Title Description

Given a double base type and floating-point type int integer exponent. Seeking exponent of the power base.

Ensure the base and exponent are not simultaneously 0

public class Solution {
    public double Power(double base, int exponent) {
        
  }
}

 

Ideas: recursion.

Because (x * x) n / 2 by solving recursive, recursive and n are each reduced by half, so the overall time complexity of the algorithm is O (logN).

achieve:

public class Solution {
    public double Power(double base, int exponent) {
        if (exponent == 0)
            return 1;
        if (exponent == 1)
            return base;
        boolean isNegative = false;
        if (exponent < 0){
            exponent = -exponent;
            isNegative = true;
        }
        double pow = Power(base * base, exponent / 2);
        if (exponent % 2 != 0)
            pow = pow * base;
        return isNegative ? 1 / pow : pow;
  }
}