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.

 1     public int thirdMax(int[] nums) {
 2         Integer max1 = null;
 3         Integer max2 = null
 
;
 4         Integer max3 = null;
 5         for (Integer n:nums)
 6         {
 7             if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue;
 8             if (max1 == null || n > max1)
 9             {
10                 max3 = max2;
11                 max2 = max1;
12                 max1 = n;
 
13             }else if (max2 == null || n > max2)
14             {
15                 max3 = max2;  
16                 max2 = n;
17             }else if (max3 == null || n > max3)
18             {
19                 max3 = n;
20             }
21         }
22         return max3 == null?max1:max3;        
23     }

414. Third Maximum Number