LeetCode: Single Number 2

Given an array of integers, every element appears three times except for one. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

public class Solution {
    public int singleNumber(int[] A) {
        int N=64;
        int countbit[]=new int[N];
        int countbitN[]=new int[N];
        int res=0;
        int resN=0;
        for(int i=0;i<A.length;i++)
        {
            int n=A[i];
            if(n>=0)
                for(int j=0;j<N;j++)
                    countbit[j]+=((n&(1<<j))>0)?1:0;
            else
                for(int j=0;j<N;j++)
                    countbitN[j]+=((-(1+n)&(1<<j))>0)?1:0;
           
        }
        for(int j=0;j<64;j++)
        {
            if(countbit[j]%3>0)res|=(1<<j);
            if(countbitN[j]%3>0)resN|=(1<<j);
        }
       
        return (resN>0)?-(resN+1):res;
    }
}