414. 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.

 

 Solution 1: use three variants first,second,third to represent the  first,second,third maximum. Note that use long long type to solve the condition that INT_MIN exists in the array. (Line 4)

 

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        long long first=LLONG_MIN, second=LLONG_MIN, third=LLONG_MIN; //
        for (int num:nums) {
            if (num>first) {
                third=second;
                second=first;
                first=num;
            }
            else if (num<first && num>second){
                third=second;
                second=num;
            }
            else if (num<second && num>third){
                third=num;
            }
        }
        return (third==LLONG_MIN)?first:third;
    }
};

Solution 2: use the functions of set: ordering and including unique key. Set is implemented with red-black tree, the complexity of operation of insert  is O(log3) and erase is O(1), so the total complexity is O(n). 

class Solution {
public:
    int thirdMax(vector<int>& nums) {
        set<int> s;
        for (int num:nums){
            s.insert(num);
            if (s.size()>3){
                s.erase(s.begin());
            }
        }
        return (s.size()<3)?*(s.rbegin()):*s.begin();
    }
};