380python
690数组
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Share 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).ide
Example 1: Input: [3, 2, 1]ui
Output: 1this
Explanation: The third maximum is 1. Example 2: Input: [1, 2]spa
Output: 2code
Explanation: The third maximum does not exist, so the maximum (2) is returned instead. Example 3: Input: [2, 2, 3, 1]排序
Output: 1it
Explanation: Note that the third maximum here means the third maximum distinct number. Both numbers with value 2 are both considered as second maximum.
思路:集合过滤重复数字,而后转换为数组,排序,大于等于三个,输出第三个,不然输出第一个元素 代码:python3
class Solution:
def thirdMax(self, nums):
s = set(nums)
arr=list(s)
arr.sort(reverse=True)
if len(arr)>=3:
return arr[2]
else:
return arr[0]
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