Pow(x n) LeetCode Solution

Here, We see Pow(x, n) problem Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc., with different approaches.

Pow(x, n) LeetCode Solution

Problem Statement ->

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

• C++ Solution
• Java Solution
• JavaScript Solution
• Python Solution

Pow(x n) Leetcode Solution C++ ->

class Solution {
public:
    double myPow(double x, int n) {
        double res = 1;
        while (n) {
            if (n % 2) res = n > 0 ? res * x : res / x;
            x = x * x;
            n /= 2;
        }
        return res;    
    }
};
Code language: C++ (cpp)

Pow(x n) Leetcode Solution Java ->

class Solution {
    public double myPow(double x, int n) {
        if(n < 0){
            n = -n;
            x = 1 / x;
        }
        double pow = 1;
        while(n != 0){
            if((n & 1) != 0){
                pow *= x;
            }     
            x *= x;
            n >>>= 1;
        }
        return pow;        
    }
}
Code language: Java (java)

Pow(x n) Leetcode Solution JavaScript ->

var myPow = function(x, n) {
    if (n===0) return 1;
    let pow = Math.abs(n);
	let result = pow%2===0 ? myPow(x*x,pow/2) : myPow(x*x,(pow-1)/2) * x;
    return n < 0 ? 1/result : result;    
};
Code language: JavaScript (javascript)

Pow(x n) Leetcode Solution Python ->

class Solution(object):
    def myPow(self, x, n):
        m = abs(n)
        ans = 1.0
        while m:
            if m & 1:
                ans *= x
            x *= x
            m >>= 1
        return ans if n >= 0 else 1 / ans
Code language: Python (python)