414.third-maximum-number

Statement

简体中文English

Metadata

• Link: 第三大的数
• Difficulty: Easy
• Tag: 数组 排序

给你一个非空数组，返回此数组中 第三大的数 。如果不存在，则返回数组中最大的数。

 

示例 1：

输入：[3, 2, 1]
输出：1
解释：第三大的数是 1 。

示例 2：

输入：[1, 2]
输出：2
解释：第三大的数不存在, 所以返回最大的数 2 。

示例 3：

输入：[2, 2, 3, 1]
输出：1
解释：注意，要求返回第三大的数，是指在所有不同数字中排第三大的数。
此例中存在两个值为 2 的数，它们都排第二。在所有不同数字中排第三大的数为 1 。

 

提示：

• 1 <= nums.length <= 104
• -231 <= nums[i] <= 231 - 1

 

进阶：你能设计一个时间复杂度 O(n) 的解决方案吗？

Metadata

• Link: Third Maximum Number
• Difficulty: Easy
• Tag: Array Sorting

Given an integer array nums, return the third distinct maximum number in this array. If the third maximum does not exist, return the maximum number.

 

Example 1:

Input: nums = [3,2,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2.
The third distinct maximum is 1.

Example 2:

Input: nums = [1,2]
Output: 2
Explanation:
The first distinct maximum is 2.
The second distinct maximum is 1.
The third distinct maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: nums = [2,2,3,1]
Output: 1
Explanation:
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2's are counted together since they have the same value).
The third distinct maximum is 1.

 

Constraints:

• 1 <= nums.length <= 104
• -231 <= nums[i] <= 231 - 1

 

Follow up: Can you find an O(n) solution?

Solution

Python 3

from typing import List


class Solution:
    def thirdMax(self, nums: List[int]) -> int:
        LOW = -0x3f3f3f3f3f3f3f3f
        max = [LOW for i in range(3)]
        for a in nums:
            cur = a
            for i in range(3):
                if cur > max[i]:
                    cur, max[i] = max[i], cur
                elif cur == max[i]:
                    break

        if max[2] == LOW:
            return max[0]

        return max[2]

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最后更新: October 11, 2023