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?

 

思路：

假设每一个整数都是由32个bit组成的。当我们把所有的数字的相应的bit加起来mod 3 所得到的是0或者是1取决于那个single number是0还是1.

 

public class Solution {
    public int singleNumber(int[] A) {
        if(A.length<3)return -1;
        int bit=0;
        int single_number=0;
        for(int i=0;i<32;i++){
            for(int j=0; j<A.length; j++){
                bit=(bit+(A[j]>>i&1))%3;
            }
            single_number+=(bit<<i);
        }
        return single_number;
    }
}