Given an 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?
数组中只出现一次的数，其余都出现三次。不得不说，位运算真的是神奇方便快捷，感觉如果精通位运算，很多问题都不是问题哎
思路：所有的数都出现过三次，那每一位上加起来一定模3为0，若不为0，那就是只出现过一次的那个数咯
时间复杂度O(32n)，空间O(1)
class Solution { public: int singleNumber(vector<int>& nums) { int ans = 0; for (int i = 0; i < 32; ++i) { int sum = 0; for (int j = 0; j < nums.size(); ++j) { if ((nums[j]>>i) & 1 == 1) { sum++; sum %= 3; } } if (sum) ans += sum << i; } return ans; } };评论区有大神写了个写法，没看懂，先码一下
public int singleNumber(int[] A) { int ones = 0, twos = 0; for(int i = 0; i < A.length; i++){ ones = (ones ^ A[i]) & ~twos; twos = (twos ^ A[i]) & ~ones; } return ones; }