LeetCode414. 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.

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LeetCode：链接

利用变量a, b, c分别记录数组第1,2,3大的数字，遍历一次数组即可，时间复杂度O(n)，但是判断的时候必须要不能等于上一个数字，因为要不相等的三个最大数字。

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