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# Third Maximum Number

## Description

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.
```

## Solution

```cpp
class Solution {
public:
    int thirdMax(vector<int>& nums) {
        int max1 = INT_MIN, max2 = INT_MIN, max3 = INT_MIN;
        bool exists_min = false;
        for(int num : nums){
            if(num == INT_MIN) exists_min = true;
            if(num == max1 || num == max2 || num <= max3) continue;
            if(num > max1){
                max3 = max2;
                max2 = max1;
                max1 = num;
            }else if(num > max2){
                max3 = max2;
                max2 = num;
            }else{
                max3 = num;
            }
        }
        if(max2 == INT_MIN || max3 == INT_MIN && !exists_min)
            return max1;
        return max3;
    }
};
```


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