Leetcode 50 - Pow(x, n)

Implement pow(x, n).

class Solution {public:    double myPow(double x, int n) {        if(n == 0) return 1;        //注意n==INT_MIN的时候会溢出，-n仍然会等于INT_MIN        if(n<0 && n!=INT_MIN) return 1.0/myPow(x,-n);        //divide and conquer        else return n%2 ? x*myPow(x*x,n/2) : myPow(x*x,n/2);    }};