### 50. Pow(x, n)

题目:
<https://leetcode.com/problems/powx-n/>


难度:

Medium


Recursive

```python
class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        if n == 0:
            return 1
        if n < 0:
            return 1 / self.myPow(x, -n)
        if n % 2 == 0:
            return self.myPow(x*x, n/2)
        else:
            return x * self.myPow(x*x, n/2)
        
```
iterative


```python
class Solution:
    def myPow(self, x, n):
        if n < 0:
            x = 1 / x
            n = -n
        pow = 1
        while n:
            if n & 1:
                pow *= x
            x *= x
            n >>= 1
        return pow
```