Third Maximum Number
- LintCode: Second Max of Array
- Review: Heap Problems
- Review: Problems With Many Details
- Tag: #manydetails, #heap
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.
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]