牛客网《剑指offer》之Python2.7实现：数值的整数次方

题目描述

给定一个double类型的浮点数base和int类型的整数exponent。求base的exponent次方

思路

简单粗暴：

# -*- coding:utf-8 -*-
class Solution:
    def Power(self, base, exponent):
        # write code here
        return base**exponent

快速幂？牛客网高玩方法，
参考博客：https://blog.csdn.net/hkdgjqr/article/details/5381028

# -*- coding:utf-8 -*-
class Solution:
    def Power(self, base, exponent):
        # write code here
        if base == 0:
            return 0
        if exponent == 0:
            return 1
        e = abs(exponent)
        tmp = base
        res = 1
        while(e > 0):
            #如果最后一位为1，那么给res乘上这一位的结果
            if (e & 1 == 1):
                res =res * tmp
            e = e >> 1
            tmp = tmp * tmp
        return res if exponent > 0 else 1/res