LeetCode_Pow(x, n)

Posted on | In | Views:

Pow(x, n)

Implement pow(x, n), which calculates x raised to the power n (x^n).
(指数计算)

Note:

  1. -100.0 < x < 100.0
  2. n is a 32-bit signed integer, within the range [−2^31, 2^31 − 1]

Example:



1. 递归

类似于二分法的方式进行指数运算。其中比较巧妙的地方是在 abs(n) 是奇数时，无论 n 是正数还是负数，left = n // 2 一定是比较小的哪一个，因此最终的结果一定是另外乘以 x。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
class Solution:
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        if n == 0:
            return 1
        if n == 1:
            return x
        if n == -1:
            return 1/x

        left = n // 2
        re = self.myPow(x, left)

        if left * 2 == n:
            return re * re
        else:
            return re * re * x