50.pow-x-n.py

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# @lc app=leetcode.cn id=50 lang=python
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# [50] Pow(x, n)
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# https://leetcode-cn.com/problems/powx-n/description/
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# algorithms
# Medium (33.22%)
# Likes:    168
# Dislikes: 0
# Total Accepted:    30.8K
# Total Submissions: 92.6K
# Testcase Example:  '2.00000\n10'
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# 实现 pow(x, n) ，即计算 x 的 n 次幂函数。
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# 示例 1:
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# 输入: 2.00000, 10
# 输出: 1024.00000
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# 示例 2:
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# 输入: 2.10000, 3
# 输出: 9.26100
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# 示例 3:
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# 输入: 2.00000, -2
# 输出: 0.25000
# 解释: 2^-2 = 1/2^2 = 1/4 = 0.25
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# 说明:
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# -100.0 < x < 100.0
# n 是 32 位有符号整数，其数值范围是 [−2^31, 2^31 − 1] 。
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# 
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class Solution(object):
    def myPow(self, x, n):
        """
        :type x: float
        :type n: int
        :rtype: float
        """
        if n==1:return x
        if n==0: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)