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?

//java里int始终占4个字节，32位，我们外层循环遍历32次，然后内层循环记录0-31位每一位出现的次数，

//内层循环结束后将结果取余于3即为当前位的值
//时间复杂度O(32 * n), 空间复杂度O(1)
// 比方说 
//1101
//1101
//1101
//0011
//0011
//0011
//1010   这个unique的   
//----
//4340  1的出现次数  
//1010  余3的话 就是那个唯一的数！
public class Solution {
    public int singleNumber(int[] A) {
        int res=0;
		int bit;
		for(int j=0;j<32;j++){
			bit=0;
			for(int i=0;i<A.length;i++){
				if((A[i]>>j&1)==1){
					bit++;
				}
			}
			bit=bit%3;
			res+=(1<<j)*bit;
		}		
		return res; 
    }
}