Pow(x, n)

 https://leetcode.com/problems/powx-n/

 approach:

if(n%2==0)     x*x              n=n/2

if(n%2!=0)      ans=ans*x   n=n-1

if (n==0) return

solution

class Solution {
public:
    double myPow(double x, int n) {
        
        double ans=1.0;
        long long nn=n;
        if(nn<0)
        {
            nn=-1*nn;
        }
        while(nn)
        {
            if(nn%2==0){
                x=x*x;
                nn=nn/2;
            }
            else{
                ans=ans*x;
                nn=nn-1;
            }
        }
        if(n<0){
            ans=(double)1.0/(double)ans;
        }
        return ans;  
    }
};