third maximum numbers

Author: weiy

Diff for: Array/ThirdMaximumNumbers.py

@@ -0,0 +1,51 @@
+"""
+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.
+
+返回第三大的数，如果不存在则返回最大的，重复的算一个。
+https://leetcode.com/problems/third-maximum-number/description/
+
+找到第 k 大个数一般的思路有：
+1. 排序后放到数组中，排序算法使用归并和快排在理想情况下都是O(nlogn)，归并比较稳定一些。之后的索引是O(1)。
+   这种的适合并不需要插入的情况，因为每次插入的时间复杂度为 O(n)。
+
+2. 建立二叉搜索树，进阶的话红黑树或AVL树。
+   这种情况下搜索和插入在理想情况下都是O(logn)。
+
+3. 就此题来说的O(n)思路：
+   建立三个变量，first,second,third,首先确保不是None，然后挨个放数据，最后输出结果。
+
+这里直接用排序了。
+
+
+"""
+class Solution(object):
+    def thirdMax(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: int
+        """
+        nums = set(nums)
+        if len(nums) < 3:
+            return max(nums)
+        
+        return sorted(nums, reverse=True)[2]