powx-n 分治实现乘方

AC:
class Solution {
public:
    double pow(double x, int n) {
        if(x == 0 && n == 0) return 1;
        if(x == 0) return 0;
        if(n == 0) return 1;
        if(n < 0){
            return 1/x * pow(1/x, -(n+1));
        }
        // n < 32 直接求解，要不然会递归太深，真是醉了。。。
        if(n < 32){
            double tmp = 1;
            for(int i = 0; i < n; i++){
                tmp = tmp * x;               
            }
            return tmp;
        }
        double tmp = pow(x, n/2);
        if(n % 2 == 1) return tmp * tmp *x;
        return tmp * tmp;
         
    }
};

class Solution {
public:
    double pow(double x, int n) {
        if (x == 0 && n < 0)
            return 0;
        double res = powCore(x, abs(n));
        if (n < 0)
            return static_cast<double>(1 / res);
        return res;
    }
    double powCore(double x, int n)
    {
        if (n == 1)
            return x;
        if (n == 0)
            return 1;
        double resTem = powCore(x, n / 2)*powCore(x, n / 2);
        if (n % 2)
            resTem *= x;
        return resTem;
    }
};