<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) {
        int n = mat.size();
        int m = mat.front().size();
        int flip = 0;
        int dir[2][2] = {{-1, 1}, {1, -1}};
        vector<int> ans;
        for (int i = 0; i < n + m - 1; ++i) {
            int x = flip ? 0 : i;
            int y = flip ? i : 0;
            if (x >= n) {
                x = n - 1;
                y = i - (n - 1);
            } else if (y >= m) {
                y = m - 1;
                x = i - (m - 1);
            }
            while (0 <= x && x < n && 0 <= y && y < m) {
                ans.push_back(mat[x][y]);
                x += dir[flip][0];
                y += dir[flip][1];
            }
            flip = !flip;
        }
        return ans;
    }
};
21983