Walls and Gates -- LeetCode 286

[Question]

You are given a m x n 2D grid initialized with these three possible values.

1. -1 - A wall or an obstacle.
2. 0 - A gate.
3. INF - Infinity means an empty room. We use the value 231 - 1 = 2147483647 to represent INF as you may assume that the distance to a gate is less than 2147483647.

Fill each empty room with the distance to its nearest gate. If it is impossible to reach a gate, it should be filled with INF.

For example, given the 2D grid:

INF  -1  0  INF
INF INF INF  -1
INF  -1 INF  -1
  0  -1 INF INF

After running your function, the 2D grid should be:

  3  -1   0   1
  2   2   1  -1
  1  -1   2  -1
  0  -1   3   4

[Analysis]
DFS starting from each gate,  each visited room store the minimum distance. It is also possible to use BFS.

[Solution]
void walls_and_gates (vector<vector<int>>& grid) {
    for (int i=0; i<grid.size(); i++)
        for(int j=0; j<grid[0].size(); j++)
            if (grid[i][j]==0) dfs(grid, i, j, 0);
    return;
}

void dfs (vector<vector<int>>& grid, int i, int j, int dist) {
    if (i>=grid.size() || i<0 || j<0 || j>=grid[0].size() || grid[i][j]<dist) return;
    grid[i][j] = dist;
    dfs(grid, i+1, j, dist+1);
    dfs(grid, i, j+1, dist+1);
    dfs(grid, i-1, j, dist+1);
    dfs(grid, i, j-1, dist+1);
}