414. Third Maximum Number

Posted 2020-11-27 boluo007

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:
    def thirdMax(self, nums: List[int]) -> int:
        
        l = [float("-inf")] * 3
        
        for i in nums:
            if i > l[0]:
                l[2] = l[1]
                l[1] = l[0]
                l[0] = i
            elif l[0] > i > l[1]:
                l[2] = l[1]
                l[1] = i
            elif l[1] > i > l[2]:
                l[2] = i
        
        if l[2] == float("-inf"):
            return l[0]
        else:
            return l[2]