LeetCode Solution - Problem Pow(x, n)

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

Example 1:

Input: x = 2.00000, n = 10
Output: 1024.00000

Example 2:

Input: x = 2.10000, n = 3
Output: 9.26100

Example 3:

Input: x = 2.00000, n = -2
Output: 0.25000
Explanation: 2-2 = 1/22 = 1/4 = 0.25

Constraints:

• -100.0 < x < 100.0
• -231 <= n <= 231-1
• n is an integer.
• -104 <= xn <= 104

Solution:

Java:

class Solution {
    public double myPow(double x, int n) {
        double ans = 1.0;
        long num = n;
        if (n < 0) {
            num = -1 * num;
        }
        while (num > 0) {
            if (num % 2 == 0) {
                x = x * x;
                num = num / 2;
            } else {
                ans = ans * x;
                num = num - 1;
            }
        }
        if (n < 0) {
            return (double)(1.0) / (double)(ans);
        }
        return ans;
    }
}

Python:

class Solution:
    def myPow(self, x: float, n: int) -> float:
        ans = 1
        if n < 0:
            x = 1 / x
            n = -n
        
        while n:
            if n % 2:
                ans *= x
            x *= x
            n //= 2

        return ans