0050._pow(x,_n).md

50. Pow(x, n)

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/powx-n/description/

内容描述

Implement pow(x, n), 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
Note:

-100.0 < x < 100.0
n is a 32-bit signed integer, within the range [−231, 231 − 1]

解题方案

思路 1

Recursive， beats 100%

class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        if n == 0:
            return 1
        if n < 0:
            return 1 / self.myPow(x, -n)
        if n & 1:  # n 为 奇数
            return x * self.myPow(x*x, n>>1)
        else:
            return self.myPow(x*x, n>>1)

思路 2

iterative

class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        if n < 0:
            x = 1 / x
            n = -n
        res = 1
        while n:
            if n & 1:
                res *= x
            x *= x
            n >>= 1
        return res