[137. Single Number II](https://leetcode.com/problems/single-number-ii/)

### 題目描述

Given an integer array `nums` where every element appears **three times** except for one, which appears **exactly once**. *Find the single element and return it.*

You must implement a solution with a linear runtime complexity and use only constant extra space.

### 範例

**Example 1:**

Input: nums = [2,2,3,2]
Output: 3

**Example 2:**

Input: nums = [0,1,0,1,0,1,99]
Output: 99

**Constraints**:

* 1 <= `nums.length` <= 3 * 10^4^
* -2^31^ <= `nums[i]` <= 2^31^ - 1
* Each element in `nums` appears exactly **three times** except for one element which appears **once**.

### 解答
#### Javascript

function singleNumber(nums) {
  let ones = 0;
  let twos = 0;

  for (const num of nums) {
    ones = (ones ^ num) & ~twos;
    twos = (twos ^ num) & ~ones;
  }

  return ones;
}

> 看答案的，學習了，本來只會用hashmap，而且也會過ㄏㄏ，只是空間複雜度有差。
> [name=Marsgoat][time=4 July, 2023]

#### C#

public class Solution {
    public int SingleNumber(int[] nums) {
        int once = 0;
        int twice = 0;

        foreach(int num in nums) {
            once = (once ^ num) & ~twice;
            twice = (twice ^ num) & ~once;
        }

        return once;
    }
}

> 看上面的，一開始有想到 XOR，但沒想到怎麼處理出現奇數次
> [name=Jim][time=Jul 5, 2023]
### Reference

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