414. Third Maximum Number

时间：2020-07-15 01:19:01 收藏：0 阅读：57

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).

给一个数组，找到第三大的值，如果没有就返回最大值。这里的值是有并列的，

[2, 2, 3, 1]->1而不是2

class Solution(object):
    def thirdMax(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        d = {}
        for value in nums:
            d[value] = 1
        first_max = float(‘-inf‘)
        second_max = float(‘-inf‘)
        third_max = float(‘-inf‘)
        for key,value in d.items():
            if key > first_max:
                third_max = second_max
                second_max = first_max
                first_max = key
            elif key > second_max:
                third_max = second_max
                second_max = key
            elif key > third_max:
                third_max = key
        return third_max if third_max != float(‘-inf‘) else first_max

原文：https://www.cnblogs.com/whatyouthink/p/13302987.html