Majority Element II
<p>Given an integer array of size <code>n</code>, find all elements that appear more than <code>&lfloor; n/3 &rfloor;</code> times.</p>

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

<pre>
<strong>Input:</strong> nums = [3,2,3]
<strong>Output:</strong> [3]
</pre>

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

<pre>
<strong>Input:</strong> nums = [1]
<strong>Output:</strong> [1]
</pre>

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

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

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5 * 10<sup>4</sup></code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you solve the problem in linear time and in <code>O(1)</code> space?</p>
Array; Hash Table; Counting; Sorting
class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {
        int n1 = 0, n2 = 0;
        int m1 = 0, m2 = 1;
        for (int m : nums) {
            if (m == m1)
                ++n1;
            else if (m == m2)
                ++n2;
            else if (n1 == 0) {
                m1 = m;
                ++n1;
            } else if (n2 == 0) {
                m2 = m;
                ++n2;
            } else {
                --n1;
                --n2;
            }
        }
        vector<int> ans;
        if (count(nums.begin(), nums.end(), m1) > nums.size() / 3) ans.push_back(m1);
        if (count(nums.begin(), nums.end(), m2) > nums.size() / 3) ans.push_back(m2);
        return ans;
    }
};