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:

用一个 HashSet 来判断元素是否曾经出现过，用 PriorityQueue来保存当前最大的3个元素（大小设为4防止 capacity 扩充引发额外开销）

public class Solution
{
    public int thirdMax(int[] nums)
    {
        PriorityQueue<Integer> pq = new PriorityQueue<>(4);
        Set<Integer> set = new HashSet<>();
        for(int i = 0; i < nums.length; i++)
        {
            if(!set.contains(nums[i])) 
            {
                set.add(nums[i]);
                pq.offer(nums[i]);
                if (pq.size() > 3)
                    pq.poll();
            }
        }
        if(pq.size() > 2)
        {
            return pq.poll();
        }
        else
        {
            int result = 0;
            while(!pq.isEmpty())
            {
                result = pq.poll();
            }
            return result;
        }
    }
}