# 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.

# List of all 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```

## 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)```
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