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.

class Solution(object):
    def thirdMax(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        a,b,c = -2147483649,-2147483649,-2147483649
        for i in range(len(nums)):
            if nums[i]>a:
                c = b
                b = a
                a = nums[i]
            elif (nums[i]>b and nums[i]<a):
                c = b
                b = nums[i]
            elif (nums[i]>c and nums[i]<b):
                c = nums[i]
        return a if (c==-2147483649 or c==b) else c