LeetCode

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;}