414. Third Maximum Number

Problem:

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. </pre> </p>

Solutions:

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        int count = 0;
        int first = INT_MIN;
        int second = INT_MIN;
        int third = INT_MIN;

        for (auto& num : nums) {
            if ((count >= 1 && num == first) || (count >= 2 && num == second) || (count >= 3 && num == third))  continue;
            ++count;
            if (num >= first) {
                third = second;
                second = first;
                first = num;
            } else if (num >= second) {
                third = second;
                second = num;
            } else if (num >= third) {
                third = num;
            }
        }

        return count >= 3 ? third : first;
    }
};

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