算法描述：

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

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

Example 1:

Input: [2,2,3,2]
Output: 3

Example 2:

Input: [0,1,0,1,0,1,99]
Output: 99

解題思路：用不進位加法模擬異或運算。這個題可以用累加和模3：由於數據要麽出現三次，要麽出現一次，所以，模3的結果只有兩種，0或者1。由模後的結果重新構成的數字就是出現一次的數字。

    int singleNumber(vector<int>& nums) {
        if(nums.size()== 1) return nums[0];
        int res = 0;
        for(int i=0; i< 32; i++){
            int sum = 0;
            
 
for(int j=0; j < nums.size(); j++){
                sum += (nums[j] >> i) & 1;
                sum %= 3;
            }
            res = res | (sum << i);
        }
        return res;
    }

LeetCode-137-Single Number II