<p>Given an integer array <code>nums</code>, return <em>the <strong>third distinct maximum</strong> number in this array. If the third maximum does not exist, return the <strong>maximum</strong> number</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

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

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2]
<strong>Output:</strong> 2
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,3,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>
The first distinct maximum is 3.
The second distinct maximum is 2 (both 2&#39;s are counted together since they have the same value).
The third distinct maximum is 1.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<strong>Follow up:</strong> Can you find an <code>O(n)</code> solution?
class Solution:
    def thirdMax(self, nums: List[int]) -> int:
        m1 = m2 = m3 = -inf
        for num in nums:
            if num in [m1, m2, m3]:
                continue
            if num > m1:
                m3, m2, m1 = m2, m1, num
            elif num > m2:
                m3, m2 = m2, num
            elif num > m3:
                m3 = num
        return m3 if m3 != -inf else m1
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