# Pow(x, n) Leetcode Solution

In this post, we are going 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.

## 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`

Now, lets see the leetcode solution of Pow(x, n) Leetcode Solution.

### Pow(x, n) Leetcode Solution 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 Solutionin 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 Solution 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 Chase2learn This tutorial is only for Educational and Learning purposes.

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