[leetcode] 137. 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?

解法一：

用一个INT数字，记录每个bit上1出现的次数。然后求次数对3的余数，就是出现一次的数字在bit上的值。

class Solution {public:    int singleNumber(vector<int>& nums) {        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;            }            res |= (sum%3)<<i;        }        return res;            }};