LeetCode 137. Single Number II

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本文介绍了一种线性时间复杂度的算法来找出数组中仅出现一次的元素，而其他元素均出现三次。通过位操作实现，避免了额外内存使用。

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?

// time: O(32n)
class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int n = nums.size();
        int result = 0;
        int bits_num = 8 * sizeof(int);
        for (int i = 0; i < bits_num; ++i) {
            int count = 0;
            int mask = 1 << i;
            for (int j = 0; j < n; ++j) {
                if (mask & nums[j]) {
                    ++count;
                }
            }
            if (count % 3) {
                result |= mask;
            }
        }
        return result;
    }
};