Last updated on October 10th, 2024 at 02:13 am

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

*List of all LeetCode Solution*

*List of all LeetCode Solution*

## Topics

Math, Recursion

## Companies

Bloomberg, Facebook, Google, Linkedin

## Level of Question

Medium

**Pow(x n) LeetCode Solution**

## Table of Contents

**Problem Statement**

Implement ** pow(x, n)**, which calculates

*raised to the power*

**x***(i.e., x*

**n**^{n}).

Example 1:Input:x = 2.00000, n = 10Output:1024.00000Example 2:Input:x = 2.10000, n = 3Output:9.26100

Example 3:Input:x = 2.00000, n = -2Output:0.25000Explanation:2-2 = 1/22 = 1/4 = 0.25

**1. 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; } };

**2. 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; } }

**3. 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; };

**4. 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