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