# Pow(x, n) Leetcode Solution In this post, you will know how to solve the Pow(x, n) Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

One more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.

## Problem

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`
• `n` is an integer.
• `-104 <= xn <= 104`

### Pow(x, n) Leetcode Solutions in Python

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

### Pow(x, n) Leetcode Solutions in CPP

```class Solution {
public:
double myPow(double x, long n) {
if (n == 0)
return 1;
if (n < 0)
return 1 / myPow(x, -n);
if (n & 1)
return x * myPow(x, n - 1);
return myPow(x * x, n / 2);
}
};
```

### Pow(x, n) Leetcode Solutions in Java

```class Solution {
public double myPow(double x, long n) {
if (n == 0)
return 1;
if (n < 0)
return 1 / myPow(x, -n);
if (n % 2 == 1)
return x * myPow(x, n - 1);
return myPow(x * x, n / 2);
}
}
```

Note: This problem Pow(x, n) is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.