The Maze II
<p>There is a ball in a <code>maze</code> with empty spaces (represented as <code>0</code>) and walls (represented as <code>1</code>). The ball can go through the empty spaces by rolling <strong>up, down, left or right</strong>, but it won&#39;t stop rolling until hitting a wall. When the ball stops, it could choose the next direction.</p>

<p>Given the <code>m x n</code> <code>maze</code>, the ball&#39;s <code>start</code> position and the <code>destination</code>, where <code>start = [start<sub>row</sub>, start<sub>col</sub>]</code> and <code>destination = [destination<sub>row</sub>, destination<sub>col</sub>]</code>, return <em>the shortest <strong>distance</strong> for the ball to stop at the destination</em>. If the ball cannot stop at <code>destination</code>, return <code>-1</code>.</p>

<p>The <strong>distance</strong> is the number of <strong>empty spaces</strong> traveled by the ball from the start position (excluded) to the destination (included).</p>

<p>You may assume that <strong>the borders of the maze are all walls</strong> (see examples).</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0505.The%20Maze%20II/images/maze1-1-grid.jpg" style="width: 573px; height: 573px;" />
<pre>
<strong>Input:</strong> maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [4,4]
<strong>Output:</strong> 12
<strong>Explanation:</strong> One possible way is : left -&gt; down -&gt; left -&gt; down -&gt; right -&gt; down -&gt; right.
The length of the path is 1 + 1 + 3 + 1 + 2 + 2 + 2 = 12.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0505.The%20Maze%20II/images/maze1-2-grid.jpg" style="width: 573px; height: 573px;" />
<pre>
<strong>Input:</strong> maze = [[0,0,1,0,0],[0,0,0,0,0],[0,0,0,1,0],[1,1,0,1,1],[0,0,0,0,0]], start = [0,4], destination = [3,2]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There is no way for the ball to stop at the destination. Notice that you can pass through the destination but you cannot stop there.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> maze = [[0,0,0,0,0],[1,1,0,0,1],[0,0,0,0,0],[0,1,0,0,1],[0,1,0,0,0]], start = [4,3], destination = [0,1]
<strong>Output:</strong> -1
</pre>

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

<ul>
	<li><code>m == maze.length</code></li>
	<li><code>n == maze[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>maze[i][j]</code> is <code>0</code> or <code>1</code>.</li>
	<li><code>start.length == 2</code></li>
	<li><code>destination.length == 2</code></li>
	<li><code>0 &lt;= start<sub>row</sub>, destination<sub>row</sub> &lt; m</code></li>
	<li><code>0 &lt;= start<sub>col</sub>, destination<sub>col</sub> &lt; n</code></li>
	<li>Both the ball and the destination exist in an empty space, and they will not be in the same position initially.</li>
	<li>The maze contains <strong>at least 2 empty spaces</strong>.</li>
</ul>
Depth-First Search; Breadth-First Search; Graph; Array; Matrix; Shortest Path; Heap (Priority Queue)
class Solution {
public:
    int shortestDistance(vector<vector<int>>& maze, vector<int>& start, vector<int>& destination) {
        int m = maze.size(), n = maze[0].size();
        int dist[m][n];
        memset(dist, 0x3f, sizeof(dist));
        int si = start[0], sj = start[1];
        int di = destination[0], dj = destination[1];
        dist[si][sj] = 0;
        queue<pair<int, int>> q;
        q.emplace(si, sj);
        int dirs[5] = {-1, 0, 1, 0, -1};
        while (!q.empty()) {
            auto [i, j] = q.front();
            q.pop();
            for (int d = 0; d < 4; ++d) {
                int x = i, y = j, k = dist[i][j];
                int a = dirs[d], b = dirs[d + 1];
                while (x + a >= 0 && x + a < m && y + b >= 0 && y + b < n && maze[x + a][y + b] == 0) {
                    x += a;
                    y += b;
                    ++k;
                }
                if (k < dist[x][y]) {
                    dist[x][y] = k;
                    q.emplace(x, y);
                }
            }
        }
        return dist[di][dj] == 0x3f3f3f3f ? -1 : dist[di][dj];
    }
};