LeetCode-137-Single Number II

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

解题思路：用不进位加法模拟异或运算。这个题可以用累加和模3：由于数据要么出现三次，要么出现一次，所以，模3的结果只有两种，0或者1。由模后的结果重新构成的数字就是出现一次的数字。

    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

原文：https://www.cnblogs.com/nobodywang/p/10353595.html