Lucky Numbers in a Matrix
<p>Given an <code>m x n</code> matrix of <strong>distinct </strong>numbers, return <em>all <strong>lucky numbers</strong> in the matrix in <strong>any </strong>order</em>.</p>

<p>A <strong>lucky number</strong> is an element of the matrix such that it is the minimum element in its row and maximum in its column.</p>

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

<pre>
<strong>Input:</strong> matrix = [[3,7,8],[9,11,13],[15,16,17]]
<strong>Output:</strong> [15]
<strong>Explanation:</strong> 15 is the only lucky number since it is the minimum in its row and the maximum in its column.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[1,10,4,2],[9,3,8,7],[15,16,17,12]]
<strong>Output:</strong> [12]
<strong>Explanation:</strong> 12 is the only lucky number since it is the minimum in its row and the maximum in its column.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[7,8],[1,2]]
<strong>Output:</strong> [7]
<strong>Explanation:</strong> 7 is the only lucky number since it is the minimum in its row and the maximum in its column.
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat[i].length</code></li>
	<li><code>1 &lt;= n, m &lt;= 50</code></li>
	<li><code>1 &lt;= matrix[i][j] &lt;= 10<sup>5</sup></code>.</li>
	<li>All elements in the matrix are distinct.</li>
</ul>
Array; Matrix
class Solution {
public:
    vector<int> luckyNumbers(vector<vector<int>>& matrix) {
        int m = matrix.size(), n = matrix[0].size();
        int rows[m];
        int cols[n];
        memset(rows, 0x3f, sizeof(rows));
        memset(cols, 0, sizeof(cols));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                rows[i] = min(rows[i], matrix[i][j]);
                cols[j] = max(cols[j], matrix[i][j]);
            }
        }
        vector<int> ans;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (rows[i] == cols[j]) {
                    ans.push_back(rows[i]);
                }
            }
        }
        return ans;
    }
};