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
"""
s = set(nums)
if(len(s) > 2):
s.remove(max(s))
s.remove(max(s))
return max(s)
else:
return max(s)
时间: 2024-10-13 23:00:25