<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>>& matrix) {
    const int m = matrix.size();
    const int n = matrix[0].size();
    vector<int> ans(m * n);
    int d = 1;  // left-bottom -> right-top
    int row = 0;
    int col = 0;

    for (int i = 0; i < m * n; ++i) {
      ans[i] = matrix[row][col];
      row -= d;
      col += d;
      // out-of-bounds
      if (row == m)
        row = m - 1, col += 2, d = -d;
      if (col == n)
        col = n - 1, row += 2, d = -d;
      if (row < 0)
        row = 0, d = -d;
      if (col < 0)
        col = 0, d = -d;
    }

    return ans;
  }
};
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