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# Pow(x, n) LeetCode Solution

## Problem – Pow(x, n) LeetCode Solution

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`
• `-104 <= xn <= 104`

### Pow(x, n) LeetCode Solution in Java

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

### Pow(x, n) LeetCode Solution in Python

``````class Solution:
def myPow(self, x, n):
if abs(x) < 1e-40: return 0
if n == 0: return 1
if n < 0: return self.myPow(1/x, -n)
lower = self.myPow(x, n//2)
if n % 2 == 0: return lower*lower
if n % 2 == 1: return lower*lower*x
``````

### Pow(x, n) LeetCode Solution in C++

``````   double pow(double x, int n) {
if (n==0) return 1;
double t = pow(x,n/2);
if (n%2) {
return n<0 ? 1/x*t*t : x*t*t;
} else {
return t*t;
}
}
``````
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