Leetcode50-Pow(x, n)(Python3)

50. Pow(x, n)

Implement pow(x, n).

My Solution

class Solution(object):
    myPow = pow

Reference (转)

1

class Solution:
    def myPow(self, x, n):
        return x ** n

2

class Solution:
    def myPow(self, x, n):
        if not n:
            return 1
        if n < 0:
            return 1 / self.myPow(x, -n)
        if n % 2:
            return x * self.myPow(x, n-1)
        return self.myPow(x*x, n/2)

3

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