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):
         """
         :type nums: List[int]
         :rtype: int
         """
         max3 = [nums[0], None, None]
         for i in range(1, len(nums)):
             if nums[i] in max3: continue
             if nums[i] > max3[0]:
                 max3[2] = max3[1]
                 max3[1] = max3[0]
                 max3[0] = nums[i]
             elif max3[1] is None or nums[i] > max3[1]:
                 max3[2] = max3[1]
                 max3[1] = nums[i]
             elif max3[2] is None or nums[i] > max3[2]:
                 max3[2] = nums[i]
             else: continue
         if None in max3: return max3[0]
         else: return max3[2]
                 

class Solution:
     def thirdMax(self, nums):
         """
         :type nums: List[int]
         :rtype: int
         """
         max3 = [nums[0], None, None]
         for i in range(1, len(nums)):
             if nums[i] in max3: continue
             if nums[i] > max3[0]:
                 max3[2] = max3[1]
                 max3[1] = max3[0]
                 max3[0] = nums[i]
             elif max3[1] is None or nums[i] > max3[1]:
                 max3[2] = max3[1]
                 max3[1] = nums[i]
             elif max3[2] is None or nums[i] > max3[2]:
                 max3[2] = nums[i]
             else: continue
         if None in max3: return max3[0]
         else: return max3[2]
                 
