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

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
public class Solution {
public int thirdMax(int[] nums) {
Integer max1 = null;
Integer max2 = null;
Integer max3 = null;
for (Integer n : nums) {
if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue;
if (max1 == null || n > max1) {
max3 = max2;
max2 = max1;
max1 = n;
} else if (max2 == null || n > max2) {
max3 = max2;
max2 = n;
} else if (max3 == null || n > max3) {
max3 = n;
}
}
return max3 == null