Longest Line of Consecutive One in Matrix
<p>Given an <code>m x n</code> binary matrix <code>mat</code>, return <em>the length of the longest line of consecutive one in the matrix</em>.</p>

<p>The line could be horizontal, vertical, diagonal, or anti-diagonal.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0562.Longest%20Line%20of%20Consecutive%20One%20in%20Matrix/images/long1-grid.jpg" style="width: 333px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[0,1,1,0],[0,1,1,0],[0,0,0,1]]
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0562.Longest%20Line%20of%20Consecutive%20One%20in%20Matrix/images/long2-grid.jpg" style="width: 333px; height: 253px;" />
<pre>
<strong>Input:</strong> mat = [[1,1,1,1],[0,1,1,0],[0,0,0,1]]
<strong>Output:</strong> 4
</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;= m, n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>4</sup></code></li>
	<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>
Array; Dynamic Programming; Matrix
class Solution {
public:
    int longestLine(vector<vector<int>>& mat) {
        int m = mat.size(), n = mat[0].size();
        vector<vector<int>> a(m + 2, vector<int>(n + 2));
        vector<vector<int>> b(m + 2, vector<int>(n + 2));
        vector<vector<int>> c(m + 2, vector<int>(n + 2));
        vector<vector<int>> d(m + 2, vector<int>(n + 2));
        int ans = 0;
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (mat[i - 1][j - 1]) {
                    a[i][j] = a[i - 1][j] + 1;
                    b[i][j] = b[i][j - 1] + 1;
                    c[i][j] = c[i - 1][j - 1] + 1;
                    d[i][j] = d[i - 1][j + 1] + 1;
                    ans = max(ans, max(a[i][j], max(b[i][j], max(c[i][j], d[i][j]))));
                }
            }
        }
        return ans;
    }
};