版权声明:版权归个人所有,未经博主允许,禁止转载 https://blog.csdn.net/danspace1/article/details/88949636
原题
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.
解法
将nums转化为集合, 然后排序求解
代码
class Solution(object):
def thirdMax(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
nums_set = set(nums)
if len(nums_set) < 3:
return max(nums_set)
return sorted(nums_set)[-3]