50\. Pow(x, n)
Medium
1636
3162
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Implement pow(
[x](http://www.cplusplus.com/reference/valarray/pow/)
[,
n
)](www.cplusplus.com/reference/v…), 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
class Solution:
def myPow(self, x: float, n: int) -> float:
neg = True if n < 0 else False
n = abs(n)
def fastPow(x, n):
if n == 0:
return 1
if n == 1:
return x
half = fastPow(x, n // 2)
if n %2:
return half * half * x
else:
return half * half
p = fastPow(x, n)
return 1.0 / p if neg else p
