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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 <= 10`

^{4}`-2`

^{31}<= nums[i] <= 2^{31}- 1

```
public int thirdMax(int[] nums) {
Integer max1 = null;
Integer max2 = null;
Integer max3 = null;
for (Integer n : nums) {
if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue;
if (max1 == null || n > max1) {
max3 = max2;
max2 = max1;
max1 = n;
} else if (max2 == null || n > max2) {
max3 = max2;
max2 = n;
} else if (max3 == null || n > max3) {
max3 = n;
}
}
return max3 == null ? max1 : max3;
}
```

```
class Solution(object):
def thirdMax(self, nums):
v = [float('-inf'), float('-inf'), float('-inf')]
for num in nums:
if num not in v:
if num > v[0]: v = [num, v[0], v[1]]
elif num > v[1]: v = [v[0], num, v[1]]
elif num > v[2]: v = [v[0], v[1], num]
return max(nums) if float('-inf') in v else v[2]
```

```
int thirdMax(vector<int>& nums) {
set<int> top3;
for (int num : nums) {
top3.insert(num);
if (top3.size() > 3)
top3.erase(top3.begin());
}
return top3.size() == 3 ? *top3.begin() : *top3.rbegin();
}
```

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