LeetCode 50. Pow(x, n)

class Solution {
public:
    double pow[35];
    double myPow(double x, int n) {
        
        if(n==0)
            return 1;
        
        long long int n2;
        int tag=0;
        
        if(n<0){
            
            n2 = (long long int)n*-1;
            tag=1;
        }
        else
            n2=n;
        
        long long int pos = 1;
        int i=0;
        double ans = x;
        pow[i]=ans;
        
        while((pos<<1)<=n2 && (pos<<1)>0)
        {
            ans*=ans;
            
            pos <<= 1;
            i++;
            
            pow[i]=ans;
        }
        
        ans=1;
        while(n2)
        {
            if(n2<pos)
            {
                pos>>=1;
                i--;
                continue;
            }
      
            n2-=pos;
            
            ans*=pow[i];
            pos>>=1;
            i--;
        }
        
        
        
        if(tag==1)
            return 1.0/ans;
        else
            return ans;
    }
};