414. Third Maximum Number

题目：

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：

本题主要是集合函数set的使用，但是有个小细节要注意，当set中个数小于3时，直接返回最大值就好了。

代码：

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        set<int> num;
        for(int i =0;i<nums.size();i++)
            num.insert(nums[i]);
        set<int>::iterator itr = --num.end();
        for(int j =num.size()-1;j>=0;j--,itr--)
            if(j==num.size()-3)
                return *itr;       
        itr = --num.end();
        return *itr;       
    }
};