Leetcode 50 Pow(x, n)

Implement pow(x, n).

快速幂，要注意n为INT_MIN和负数的情况，细碎小恶魔

class Solution {public:    double myPow(double x, long long n) {        double result=1,base=x;        while(n)        {            if(n<0)            {                base=1/base;                n=-n;            }            if(n&1) result*=base;            base*=base;            n>>=1;        }        return result;    }};