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: 1Explanation: The third maximum is 1.

Example 2:

Input: [1, 2]Output: 2Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: [2, 2, 3, 1]Output: 1Explanation: Note that the third maximum here means the third maximum distinct number.Both numbers with value 2 are both considered as second maximum.

这道题的测试用例有MIN_VALUE 这个值，因此如果想用三个变量保存前三大的值的话，这三个变量得是long型的，有点蛋疼；于是用priorityqueue做可以避免这些事情；但是这道题是不允许重复的，所以要用一个set保存结果。这样扫一遍就知道第三大是少多了。如果不存在第三大，那么要么是1 要么是2， 2的话再删除一个元素，返回最大值。

代码：

public int thirdMax(int[] nums) {        Queue<Integer> queue = new PriorityQueue<>();        Set<Integer> set = new HashSet<>();        for(int num: nums){            if(!set.contains(num)){                set.add(num);                queue.offer(num);                if(queue.size()>3) queue.poll();            }        }        if(queue.size()==2) queue.poll();        return queue.peek();    }