剑指offer:数值的整数次方

直接上快速幂
原题链接：https://www.acwing.com/problem/content/description/26/

class Solution {
    
public:
    double Power(double base, int exponent) {
    
        typedef long long LL;
        bool is_minus = exponent < 0;
        double res = 1;
        
        for(LL k = abs(LL(exponent)); k; k >>= 1)
        {
    
            if(k & 1) res *= base;
            base *= base;
        }
        if(is_minus) res = 1 / res;
        return res;
    }
};

class Solution(object):
    def Power(self, base, exponent):
        """ :type base: float :type exponent: int :rtype: float """
        is_minus = exponent < 0
        res = 1
        
        k = abs(exponent)
        while k:
            if k & 1:
                res *= base
            base = base * base
            k >>= 1
        
        if is_minus:
            res = 1 / res
        
        return res