原题是：
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.
思路是：
利用set,list转换去除list里的重复元素。
然后用List的排序。
代码是：
class Solution:
def thirdMax(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums = list(set(nums))
nums.sort()
if len(nums) >2:
return nums[-3]
else :
return nums[-1]
