# 414. Third-Maximum-Number

difficulty: Easy

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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.
```

## Method One

```
class Solution {
    public int thirdMax(int[] nums) {
        Set<Integer> set = new HashSet<>();
        
        int first = Integer.MIN_VALUE;
        int second = Integer.MIN_VALUE;
        int third = Integer.MIN_VALUE;
        for(int n : nums){
            if( set.contains(n)){
                continue;
            }
            if( n > first){
                third = second;
                second = first;
                first = n;
            }else if ( n > second ){
                third = second;
                second = n;
            }else if ( n > third ){
                third = n;
            }
            set.add(n);
        }
        return set.size() >= 3 ? third : first;
    }
}
​
```
