Third Maximum Number @LeetCode

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.

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SIMPLE c++ IMPLEMENTATION :
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class Solution {
public:
    int thirdMax(vector<int>& nums) {
        if(nums.size()==1)
            return nums[0] ;
        else if(nums.size()==2)
            return nums[0]>nums[1]?nums[0]:nums[1] ;
       
        long first,second,third ;
        first= (long)INT_MIN-1;
        second= (long)INT_MIN-1 ;
        third= (long)INT_MIN-1 ;
       
        for(int i=0;i<nums.size();i++){
           
            if(nums[i]==first||(nums[i]==second||nums[i]==third))
                continue ;
           
            if(nums[i]>=first){
                third=second ;
                second=first ;
                first=nums[i] ;
            }
           
            else if(nums[i]>=second){
                third=second ;
                second=nums[i] ;
            }
           
            else if(nums[i]>=third)
                third=nums[i] ;
        }
       
        return (third== (long)INT_MIN-1)?(first>second?first:second):third ;
    }
};

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