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[0] if len(l) != 3 else l[-1]