Given integer array nums, return the third maximum number in this array. If the third maximum does not exist, return the maximum number.

Example 1:
Input: nums = [3,2,1]
Output: 1
Explanation: The third maximum is 1.

Example 2:
Input: nums = [1,2]
Output: 2
Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:
Input: nums = [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.

Constraints:****

  • 1 <= nums.length <= 104
  • -2^31 <= nums[i] <= 2^31 – 1

Follow up: Can you find an O(n) solution?

Solution in python:

class Solution:
    def thirdMax(self, nums: List[int]) -> int:
        result = []
        for i in range(3):
            max_num = float('-inf')
            for item in nums:
                if (item > max_num) and (item not in result):
                    max_num = item
            if max_num != float('-inf'):
                result.append(max_num)
        if len(result) == 3:
            return result[2]
        else: return result[0]
最后修改日期: 2021年1月29日

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