# [414. Third Maximum Number](https://leetcode.com/problems/third-maximum-number/)

## 题目

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.


## 题目大意

给定一个非空数组，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。要求算法时间复杂度必须是 O(n)。


## 解题思路

- 水题，动态维护 3 个最大值即可。注意数组中有重复数据的情况。如果只有 2 个数或者 1 个数，则返回 2 个数中的最大值即可。

## 代码

```go

package leetcode

import (
	"math"
)

func thirdMax(nums []int) int {
	a, b, c := math.MinInt64, math.MinInt64, math.MinInt64
	for _, v := range nums {
		if v > a {
			c = b
			b = a
			a = v
		} else if v < a && v > b {
			c = b
			b = v
		} else if v < b && v > c {
			c = v
		}
	}
	if c == math.MinInt64 {
		return a
	}
	return c
}

```