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.
考虑到存在相同的数的情况，用set消除相同的数。然后再找到第三大的数。
class Solution(object): def thirdMax(self, nums): """ :type nums: List[int] :rtype: int """ nums =set(nums) if len(nums) < 3: return max(nums) nums.remove(max(nums)) nums.remove(max(nums)) return max(nums)