# 414.third-maximum-number

## Statement

• Difficulty: Easy
• Tag: `数组` `排序`

``````输入：[3, 2, 1]

``````输入：[1, 2]

``````

``````输入：[2, 2, 3, 1]

• `1 <= nums.length <= 104`
• `-231 <= nums[i] <= 231 - 1`

Given an integer array `nums`, return the third distinct 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 first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.
``````

Example 2:

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

Example 3:

``````Input: nums = [2,2,3,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2's are counted together since they have the same value).
The third distinct maximum is 1.
``````

Constraints:

• `1 <= nums.length <= 104`
• `-231 <= nums[i] <= 231 - 1`

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

## Solution

``````from typing import List

class Solution:
def thirdMax(self, nums: List[int]) -> int:
LOW = -0x3f3f3f3f3f3f3f3f
max = [LOW for i in range(3)]
for a in nums:
cur = a
for i in range(3):
if cur > max[i]:
cur, max[i] = max[i], cur
elif cur == max[i]:
break

if max[2] == LOW:
return max[0]

return max[2]
``````