[LeetCode]Pow(x,n)二分探索

class Solution {

public:

    double pow(double x, int n) {

        if(n==0)    return 1;

        bool nNeg = false;

        long long int nn = n;

        if(n<0){

            nn = - nn ;

            nNeg  =true;

        }

        bool xNeg = false;

        if(x<0){

            x = -x;

            if(n%2==1)

                xNeg = true;

        }

        double ret = 1;

        while(nn){

            if(nn&1)

                ret *= x;

            x *=x;

            nn>>=1;

        }

        if(nNeg==true)

            ret = 1/ret;

        if(xNeg==true)

            return -ret;

        return ret;

    }

};