[剑指Offer] 12.数值的整数次方

【思路1】递归

class Solution {
public:
    double Power(double base, int exponent) {
        if(exponent < 0){
            base = 1/base;
            exponent = -exponent;
        }
        if(exponent == 0){
            return 1;
        }else{
            return base * Power(base, exponent - 1);
        }
    }
};

【思路2】快速幂

class Solution {
public:
    double Power(double base, int exponent) {
        long long p = abs((long long)exponent);
        double ans = 1.0;
        while(p != 0) {
            if(p & 1) {
                ans *= base;
            }
            base *= base;
            p >>= 1;
        }
        return exponent > 0 ? ans : 1/ans;
    }
};