Leetcode 414. Third Maximum Number
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.
1 class Solution(object): 2 def thirdMax(self, nums): 3 """ 4 :type nums: List[int] 5 :rtype: int 6 """ 7 a = b = c = None 8 for x in nums: 9 if x > a: 10 a, b, c = x, a, b 11 if b<x<a: 12 a, b, c = a, x, b 13 if c<x<b: 14 a, b, c = a, b, x 15 if c == None: 16 return a 17 else: 18 return c
本题学到的是，python的判别条件里可以用 a<x<b 这种格式，方便了很多。另外就是L7里面a = b = c = None。这种赋值是可以的。但是需要注意由于Python的 = 其实是pointer。所以如果用一下的code，我们其实同时改变了a和b的值（变成了[1,2]）。以为他们是指向同一个对象的。
1 a = b = [1] 2 a.append(2)
