lintcode83- Single Number II- midium

public class Solution {
    /**
     * @param A : An integer array
     * @return : An integer 
     */
    public int singleNumberII(int[] A) {
        // write your code here
        int res = 0;
        for (int i = 0; i < 32; i++){
            int rsBit = 0;
            for (int idx = 0; idx < A.length; idx++){
                int bit = A[idx] >> i;
                bit &= 1;
                rsBit += bit;
            }
            rsBit %= 3;
            res += rsBit << i;
        }
        return res;
    }
}

public class Solution {
    /**
     * @param A : An integer array
     * @return : An integer 
     */
    public int singleNumberII(int[] A) {
        // write your code here
        int ones = 0, twos = 0, threes = 0;
        
        for(int i = 0; i < A.length; i++){
            //某数第一次进来，ones twos threes  1 0 0
            //某数第二次进来，ones twos threes  0 1 0
            //某数第三次进来，ones twos threes先1 1 1, 后 0 0 1
            //某数第四次进来，ones twos threes  1 0 0
            //循环
            twos |= ones & A[i];
            ones ^= A[i];
            threes = ones & twos;
            //第三次来时抹去一二痕迹
            twos &= ~three;
            ones &= ~three;
        }
        
        return ones;
    }
}