Implement pow(x, n), which calculates x raised to the power n (xn).
Example 1:
Input: 2.00000, 10 Output: 1024.00000 Example 2:
Input: 2.10000, 3 Output: 9.26100 Example 3:
Input: 2.00000, -2 Output: 0.25000 Explanation: 2-2 = 1/22 = 1/4 = 0.25 Note:
-100.0 < x < 100.0 n is a 32-bit signed integer, within the range [−231, 231 − 1]
My code:
class Solution:
def myPow(self, x, n):
"""
:type x: float
:type n: int
:rtype: float
"""
if(n == 0):
return 1
if n<0:
n = -n;
x = 1/x;
if n%2 ==0:
return self.myPow(x*x, n/2)
else:
return x*self.myPow(x*x,(n-1)/2)
Solution: https://leetcode.com/problems/powx-n/