一、问题描述：
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.
二、解决思路：
最先想到的但是效率不高的方法是用Treeset存放数组中的uninque元素，Treeset 可以自动排序（升序）
三、代码：
public class Solution { public int thirdMax(int[] nums) { TreeSet<Integer> set = new TreeSet<Integer>(); for (int i = 0; i < nums.length; i++) { set.add(nums[i]); } int[] tmp = new int[set.size()]; Iterator it = set.iterator(); int id = set.size() - 1; while (it.hasNext()) { tmp[id] = (int) it.next(); id--; } if (set.size() < 3) { return tmp[0]; } return tmp[2]; } }
