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.

思路：

首先想到的是用一个map存储数组中的值，作为key，因为map中的key是有序的，于是可以直接拿来用。

int thirdMax(vector<int>& nums)
{
  map<int,int>m;
  map<int,int>::iterator it;
  for(int i=0;i<nums.size();i++)
  {
    m[nums[i]]++;    
  }
  if(m.size()<3)
  {
    it = m.end();
    it--;    
  }
  else
  {
    it = m.end();
    it--,it--,it--;  
  }
  return it->first;
}

又看到大神用的set，随时保持set的size不超过3个，这样更简洁！

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();
}

参考：

https://discuss.leetcode.com/topic/63903/short-easy-c-using-set

[leetcode-414-Third Maximum Number]