[leetcode]Pow(x, n)

class Solution {
public:
double epow(double x,int n){
if(n == 0) return 1;
double tmp = epow(x , n / 2);
if(n % 2 == 1)
return tmp * tmp * x;
else return tmp * tmp;
}
double pow(double x, int n) {
if(abs(x - 0) < 1e-10) return 0;
if(n < 0) return 1/epow(x,-n);else
return epow(x,n);
}
};