LeetCode 50.Pow(x, n)

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This question is relatively simple, is to achieve an exponential function, topics are as follows:

First, how to answer this question, we might think, in many languages have built this function, you can directly call the library function, this time time complexity degree $O (1)$ , there is a way violent solution, a write cycle, the number of cycles for the $n$ , every time multiplied $x$ , so the time complexity of this time $O (N)$ , which are relatively easy way to think of.
Next we can be analyzed from another perspective, we can use a recursive method, the entire $n$ -th power divided by two, of course, necessary to determine $n$ parity until $n$ is 1, the code is as follows:

class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        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)

This recursive method, it can be exponential function recursively split, hoping to help everyone understand the recursive algorithm, thank you.