<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) {
        vector<int> res;
        size_t row = mat.size();
        size_t col = mat[0].size();
        int r, c;
        for(size_t d = 0; d < row + col - 1; d++){
            if(d%2){
                if(d < col){
                    r = 0;
                    c = d;
                }
                else{
                    r = d - col + 1;
                    c = col - 1;
                }
                while(r<row&&c>=0){
                    res.push_back(mat[r][c]);
                    r++;
                    c--;
                }
            }
            else{
                if(d < row){
                    r = d;
                    c = 0;
                }
                else{
                    r = row - 1;
                    c = d - row + 1;
                }
                while(c<col&&r>=0){
                    res.push_back(mat[r][c]);
                    c++;
                    r--;
                }
            }
        }
        return res;
    }
};
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