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
func majorityElement(nums []int) []int {
	var n1, n2 int
	m1, m2 := 0, 1
	for _, m := range nums {
		if m == m1 {
			n1++
		} else if m == m2 {
			n2++
		} else if n1 == 0 {
			m1, n1 = m, 1
		} else if n2 == 0 {
			m2, n2 = m, 1
		} else {
			n1, n2 = n1-1, n2-1
		}
	}
	n1, n2 = 0, 0
	for _, m := range nums {
		if m == m1 {
			n1++
		} else if m == m2 {
			n2++
		}
	}
	var ans []int
	if n1 > len(nums)/3 {
		ans = append(ans, m1)
	}
	if n2 > len(nums)/3 {
		ans = append(ans, m2)
	}
	return ans
}