Third Maximum Number

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

class Solution(object):
    def thirdMax(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        l = [float("-inf")] * 3
        for n in nums:
            if n not in l and n > l[0]:
                heapq.heappushpop(l, n)
                print l
        return l[0] if l[0] != float('-inf') else max(l)

public class Solution {
    public int thirdMax(int[] nums) {
        PriorityQueue<Integer> q = new PriorityQueue<Integer>();
        Set<Integer> set = new HashSet<>();

        for(int i : nums) {
            if(!set.contains(i)) {
                q.offer(i);
                set.add(i);
                if(q.size() > 3)
                    set.remove(q.poll());
            }
        }

        if(q.size() < 3) {
            while(q.size() > 1)
                q.poll();
        }

        return q.peek();
    }
}