# Leetcode: Third Maximum Number

Third Maximum Number Similar Problems:

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.
```

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/third-maximum-number
## Basic Ideas: minheap
##
## Complexity: O(n), Space O(1)
import heapq
class Solution(object):
def thirdMax(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
q = []
for num in nums:
if num not in q:
heapq.heappush(q, num)
if len(q) > 3: heapq.heappop(q)

l = []
while len(q) != 0:
l.insert(0, heapq.heappop(q))
return l if len(l) != 3 else l[-1]
```

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