Pow(x, n) LeetCode Solution

Problem – Pow(x, n) LeetCode Solution

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

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
• -104 <= xn <= 104

Pow(x, n) LeetCode Solution in Java

public class Solution {
    public double pow(double x, int n) {
        if(n == 0)
            return 1;
        if(n<0){
            n = -n;
            x = 1/x;
        }
        return (n%2 == 0) ? pow(x*x, n/2) : x*pow(x*x, n/2);
    }
}

Pow(x, n) LeetCode Solution in Python

class Solution:
    def myPow(self, x, n):
        if abs(x) < 1e-40: return 0 
        if n == 0: return 1
        if n < 0: return self.myPow(1/x, -n)
        lower = self.myPow(x, n//2)
        if n % 2 == 0: return lower*lower
        if n % 2 == 1: return lower*lower*x

Pow(x, n) LeetCode Solution in C++

   double pow(double x, int n) {
        if (n==0) return 1;
        double t = pow(x,n/2);
        if (n%2) {
            return n<0 ? 1/x*t*t : x*t*t;
        } else {
            return t*t;
        }
    }

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