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.

LeetCode：

利用變數a, b, c分別記錄陣列第1,2,3大的數字，遍歷一次陣列即可，時間複雜度O(n)，但是判斷的時候必須要不能等於上一個數字，因為要不相等的三個最大數字。

class Solution(object):
    def thirdMax(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        a = b = c = float("-inf")
        for i in nums:
            if i > a:
                a, b, c = i, a, b
            elif a > i > b:
                b, c = i, b
            elif b > i > c:
                c = i
        return c if c != float('-inf') else a