<p>Given an <code>m x n</code> matrix <code>mat</code>, return <em>an array of all the elements of the array in a diagonal order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/04/10/diag1-grid.jpg" style="width: 334px; height: 334px;" />
<pre>
<strong>Input:</strong> mat = [[1,2,3],[4,5,6],[7,8,9]]
<strong>Output:</strong> [1,2,4,7,5,3,6,8,9]
</pre>

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

<pre>
<strong>Input:</strong> mat = [[1,2],[3,4]]
<strong>Output:</strong> [1,2,3,4]
</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>-10<sup>5</sup> &lt;= mat[i][j] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    vector<int> findDiagonalOrder(vector<vector<int>>& mat) {
        const int m = mat.size();
        const int n = mat[0].size();
        vector<int> res;
        res.reserve(m * n);

        vector<vector<int>> pos;
        for (int i = 0; i < m; i++) { pos.push_back({i, 0}); }
        for (int j = 1; j < n; j++) { pos.push_back({m-1, j}); }

        bool rev = false;
        for (const auto& p : pos) {
            // start at [p[0], p[1]], i--, j++
            int start = res.size();
            for (int i = p[0], j = p[1]; i >= 0 && j < n; i--, j++) {
                res.push_back(mat[i][j]);
            }
            if (rev) {
                std::reverse(res.begin() + start, res.end());
            }
            rev = !rev;
        }

        return res;
    }
};
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