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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.

**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**.

```
class Solution:
def singleNumber(self, nums):
single = 0
for i in range(32):
count = 0
for num in nums:
if num & (1<<i) == (1<<i): count += 1
single |= (count%3) << i
return single if single < (1<<31) else single - (1<<32)
```

```
int singleNumber(vector<int>& s)
{
vector<int> t(32);////Made a array contain 32 elements.
int sz = s.size();
int i, j, n;
for (i = 0; i < sz; ++i)
{
n = s[i];
for (j = 31; j >= 0; --j)
{
t[j] += n & 1;//Find the last digit.
n >>= 1;
if (!n)
break;
}
}
int res = 0;
for (j = 31; j >= 0; --j)
{
n = t[j] % 3;//"3" represents k times.
if (n)
res += 1 << (31 - j);
}
return res;
}
```

```
public int singleNumber(int[] nums) {
int ones = 0, twos = 0, threes = 0;
for (int i = 0; i < nums.length; i++) {
// twos holds the num that appears twice
twos |= ones & nums[i];
// ones holds the num that appears once
ones ^= nums[i];
// threes holds the num that appears three times
threes = ones & twos;
// if num[i] appears three times
// doing this will clear ones and twos
ones &= ~threes;
twos &= ~threes;
}
return ones;
}
```

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