# 414. Third Maximum Number

Given a **non-empty** array of integers, return the **third** maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

**Example 1:**\\

```
Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.
```

**Example 2:**\\

```
Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
```

**Example 3:**\\

```
Input: [2, 2, 3, 1]

Output: 1

Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.
```

```cpp
int thirdMax(vector<int>& nums) { // time: O(n); space: O(1)
    long mx1 = numeric_limits<long>::min(), mx2 = numeric_limits<long>::min(), mx3 = numeric_limits<long>::min();
    for (int num : nums) {
        if (mx1 == num || mx2 == num || mx3 == num) continue;
        if (num > mx1) {
            mx3 = mx2;
            mx2 = mx1;
            mx1 = num;
        } else if (num > mx2) {
            mx3 = mx2;
            mx2 = num;
        } else if (num > mx3) {
            mx3 = num;
        }
    }
    return mx3 != numeric_limits<long>::min() ? mx3 : mx1;
}
```
