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.

这是在考容器的使用吗。。。顺便复习了一下，set，map这种容器的迭代器不能加减

 1 class Solution {
 2 public:
 3     int thirdMax(vector<int>& nums) {
 4         
 5         set<int> s(nums.begin(), nums.end());
 6         if(s.size()<3)
 7             return *s.rbegin();
 8         return *(++(++s.rbegin()));
 9         
10     }
11 };