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.

给定一非空整数数组，返回其第三小的元素，若不存在则返回最大的数。算法时间控制在O(n)。这里重复的数字视为一个。

思路
一般选择问题，选第三大。
直接进行扫描，记录第三大或最大的做法存在如下错误：当若使用int型变量记录元素，当第三大恰好为INT_MIN时将出现错误，因为无法判定这里的INT_MIN是扫描出的第三大还是没找到第三大而留下的变量初值。
正确的做法是使用set，利用set元素排列有序的特性。

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        set<int> top3;
        for (int num : nums) {
            top3.insert(num);
            if (top3.size() > 3)
                top3.erase(top3.begin());
        }
        return top3.size() == 3 ? *top3.begin() : *top3.rbegin();
    }
};