【leetcode】414. Third Maximum Number【E】【85】

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本文介绍了一种寻找数组中第三大数值的方法。通过使用set去除重复元素，并将数组排序后返回第三大元素。若不存在第三大元素，则返回最大值。

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.

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找第三大的数，刚开始还想着要不要维护一个大小为3的最大堆

后来一想，直接set一下，去一下重

然后输出第三大就行了啊

果然就行了

class Solution(object):
    def thirdMax(self, nums):
        
        nums = set(nums)
        nums = list(nums)
        
        nums.sort(reverse = True)
        #print nums
        
        if len(nums) < 3:
            return max(nums)
        
        return nums[2]