Leetcode 286. Walls and Gates

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

-1

 - A wall or an obstacle.

0

 - A gate.

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

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