<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k</code> <em>most frequent elements</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [1,1,1,2,2,3], k = 2
<strong>Output:</strong> [1,2]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [1], k = 1
<strong>Output:</strong> [1]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>k</code> is in the range <code>[1, the number of unique elements in the array]</code>.</li>
	<li>It is <strong>guaranteed</strong> that the answer is <strong>unique</strong>.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Your algorithm&#39;s time complexity must be better than <code>O(n log n)</code>, where n is the array&#39;s size.</p>

class Solution {
public:
    vector<int> topKFrequent(vector<int>& nums, int k) {
        int* H;
        vector<int> mostFrequentK;
        int max = nums[0], min = nums[0];
        for (int i = 1; i < nums.size(); i++) {
            if (max < nums[i])
                max = nums[i];
        }
        for (int i = 1; i < nums.size(); i++) {
            if (min > nums[i])
                min = nums[i];
        }
        int lengthOfHashmap = max - min + 1;
        H = new int[lengthOfHashmap]();
        for (int i = 0; i < nums.size(); i++) {
            H[nums[i] - min]++;
        }
        for (int i = 0; i < k; i++) {
            int maxVal=-1;
            int maxValIdx = -1;
            for (int j = 0; j < lengthOfHashmap; j++) {
                if(H[j]>maxVal) {
                    maxVal = H[j];
                    maxValIdx=j;
                }
            }
            mostFrequentK.push_back(maxValIdx+min);
            H[maxValIdx]=-1;
        }
        delete[] H;
        return mostFrequentK;
    }
};
16000