<p>You are given two integers <code>m</code> and <code>n</code>, which represent the dimensions of a matrix.</p>

<p>You are also given the <code>head</code> of a linked list of integers.</p>

<p>Generate an <code>m x n</code> matrix that contains the integers in the linked list presented in <strong>spiral</strong> order <strong>(clockwise)</strong>, starting from the <strong>top-left</strong> of the matrix. If there are remaining empty spaces, fill them with <code>-1</code>.</p>

<p>Return <em>the generated matrix</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/09/ex1new.jpg" style="width: 240px; height: 150px;" />
<pre>
<strong>Input:</strong> m = 3, n = 5, head = [3,0,2,6,8,1,7,9,4,2,5,5,0]
<strong>Output:</strong> [[3,0,2,6,8],[5,0,-1,-1,1],[5,2,4,9,7]]
<strong>Explanation:</strong> The diagram above shows how the values are printed in the matrix.
Note that the remaining spaces in the matrix are filled with -1.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/05/11/ex2.jpg" style="width: 221px; height: 60px;" />
<pre>
<strong>Input:</strong> m = 1, n = 4, head = [0,1,2]
<strong>Output:</strong> [[0,1,2,-1]]
<strong>Explanation:</strong> The diagram above shows how the values are printed from left to right in the matrix.
The last space in the matrix is set to -1.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m, n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li>The number of nodes in the list is in the range <code>[1, m * n]</code>.</li>
	<li><code>0 &lt;= Node.val &lt;= 1000</code></li>
</ul>


auto _ = []() { // NOLINT(bugprone-reserved-identifier, cppcoreguidelines*)
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    return 0;
}();

#define setmatrix(m, iter)                                                     \
    do { /* NOLINT(cppcoreguidelines-avoid-do-while) */                        \
        int val = -1;                                                          \
        if (iter) {                                                            \
            val = (iter)->val;                                                 \
            (iter) = (iter)->next;                                             \
        }                                                                      \
        (m) = val;                                                             \
    } while (false);

class Solution {
  public:
    vector<vector<int>> spiralMatrix(int m, int n, ListNode *head) {
        vector<vector<int>> matrix(m);
        for (int i = 0; i < m; ++i) {
            matrix[i] = vector<int>(n);
        }
        auto *iter = head;
        int ilow = 0, ihigh = m, jlow = 0, jhigh = n;
        // inclusive lower and exclusive higher limits
        for (int j = 0, i = 0;;) {
            while (j < jhigh) { // right
                setmatrix(matrix[i][j], iter);
                j++;
            }
            j--;
            i++;
            ilow++;
            if (!(ilow < ihigh)) {
                break;
            }
            while (i < ihigh) { // down
                setmatrix(matrix[i][j], iter);
                i++;
            }
            i--;
            j--;
            jhigh--;
            if (!(jlow < jhigh)) {
                break;
            }
            while (j >= jlow) { // left
                setmatrix(matrix[i][j], iter);
                j--;
            }
            j++;
            i--;
            ihigh--;
            if (!(ilow < ihigh)) {
                break;
            }
            while (i >= ilow) { // up
                setmatrix(matrix[i][j], iter);
                i--;
            }
            jlow++;
            i++;
            j++;
            if (!(jlow < jhigh)) {
                break;
            }
        }
        return matrix;
    }
};


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