Leetcode 286. Walls and Gates

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 than2147483647.

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

public class Solution {
    public void wallsAndGates(int[][] rooms) {
        if(rooms == null || rooms.length == 0) {
            return;
        }
        
        Queue<int[]> queue = new LinkedList<>();
        for(int i = 0; i < rooms.length; i++) {
            for(int j = 0; j < rooms[0].length; j++) {
                if(rooms[i][j] == 0) {
                    queue.add(new int[]{i, j});
                }
            }
        }
        while(!queue.isEmpty()) {
            int[] gate = queue.poll();
            int row = gate[0], col = gate[1];
            if(row > 0 && rooms[row - 1][col] == Integer.MAX_VALUE) {
                rooms[row - 1][col] = rooms[row][col] + 1;
                queue.add(new int[]{row - 1, col});
            }
            if(row < rooms.length - 1 && rooms[row + 1][col] == Integer.MAX_VALUE) {
                rooms[row + 1][col] = rooms[row][col] + 1;
                queue.add(new int[]{row + 1, col});
            }
            if(col > 0 && rooms[row][col - 1] == Integer.MAX_VALUE) {
                rooms[row][col - 1] = rooms[row][col] + 1;
                queue.add(new int[]{row, col - 1});
            }
            if(col < rooms[0].length - 1 && rooms[row][col + 1] == Integer.MAX_VALUE) {
                rooms[row][col + 1] = rooms[row][col] + 1;
                queue.add(new int[]{row, col + 1});
            }
        }
    }
}

This approach is to use the BFS search, this is quite reasonable, because it needs to return the maximum value to the gate position