leetcode---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.

class Solution {public:    int thirdMax(vector<int>& nums)     {        int max1 = INT_MIN;        int max2 = INT_MIN;        int max3 = INT_MIN;        set<int> s;        for(int i=0; i<nums.size(); i++)        {            if(nums[i] > max1 && !s.count(nums[i]))            {                max3 = max2;                max2 = max1;                max1 = nums[i];            }            else if(nums[i] > max2 && !s.count(nums[i]))            {                max3 = max2;                max2 = nums[i];            }            else if(nums[i] > max3 && !s.count(nums[i]))                max3 = nums[i];            s.insert(nums[i]);        }        if(s.size() < 3)        {            return max1;        }        else        {            if(max3 != max2)                return max3;            else if(max2 != max1)                return max2;            return max1;        }    }};