LeetCode-Single Number II

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?

class Solution {
public:
    void  toTribit(vector<int>& bits,int val){
        
        if(val<0){
            bits[31]=1;
            val=-val;
        }
        else bits[31]=0;
        int ind=0;
        while(val>0){
            bits[ind]=val%3;
            val/=3;
            ind++;
        }
        for(int i=ind;i<bits.size()-1;i++)bits[i]=0;
    }
    void Xor(vector<int>& bits1,vector<int>& bits2){
        for(int i=0;i<bits1.size();i++){
            bits1[i]=(bits1[i]+bits2[i])%3;
        }
    }
    int tribitToInt(vector<int>& bits){
        int ret=0;
        if(bits[31]==1){
            for(int i=bits.size()-2;i>=0;i--){
                ret*=3;
                ret-=bits[i];
            }
            if(ret==0)return 0x80000000;
        }
        else{
            for(int i=bits.size()-2;i>=0;i--){
                ret*=3;
                ret+=bits[i];
            }
        }
        return ret;
    }
    int singleNumber(int A[], int n) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<int> bits1,bits2;
        if(n<1)return 0;
        bits1.resize(32,0);
        bits2.resize(32,0);
        toTribit(bits1,A[0]);
    //    cout<<tribitToInt(bits1)<<endl;
        for(int i=1;i<n;i++){
            toTribit(bits2,A[i]);
            Xor(bits1,bits2);
        //    cout<<tribitToInt(bits2)<<endl;
        }
        if(n%3==2){
            for(int i=0;i<32;i++){
                switch(bits1[i]){
                case 0: bits1[i]=0;break;
                case 1:bits1[i]=2;break;
                case 2:bits1[i]=1;break;
                default:break;
                }
            }
        }
        return tribitToInt(bits1);
    }
};