動機
複習快速pow
Problem
Implement pow(x, n), which calculates x
raised to the power n
(i.e., xn
).
Example 1:
Input: x = 2.00000, n = 10Output: 1024.00000
Example 2:
Input: x = 2.10000, n = 3Output: 9.26100
Example 3:
Input: x = 2.00000, n = -2Output: 0.25000Explanation: 2-2 = 1/22 = 1/4 = 0.25
Constraints:
-100.0 < x < 100.0
-231 <= n <= 231-1
-104 <= xn <= 104
Sol
def dp(x,n):
if n == 0:
return 1
elif n == 1:
return x
else:
if n % 2 == 0:
half = dp(x,n//2)
return half*half
else:
return dp(x,n-1)*x
class Solution:
def myPow(self, x: float, n: int) -> float:
ret = dp(x,abs(n))
return 1/ret if n < 0 else ret