# 414. Third Maximum Number

Easy
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?

### 解題

Runtime: 5 ms, faster than 77.45%
Memory Usage: 2.9 MB, less than 72.06%
func thirdMax(nums []int) int {
max1 := math.MinInt64
max2 := math.MinInt64
max3 := math.MinInt64
for _, val := range nums {
if val > max1 {
max3 = max2
max2 = max1
max1 = val
} else if val > max2 && val!= max1 {
max3 = max2
max2 = val
} else if val > max3 && val!=max2 && val!=max1 {
max3 = val
}
}
if max3==math.MinInt64 { return max1 }
return max3
}