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

Similar:130. Surrounded Regions200. Number of Islands317. Shortest Distance from All Buildings

public class Solution {
public void wallsAndGates(int[][] rooms) {
if (rooms.length == 0 || rooms[0].length==0)
return;
Queue<int[]> queue = new LinkedList<int[]>();
int m = rooms.length;
int n = rooms[0].length;
int[] shift = {1, 0, -1, 0, 1};
for (int i=0; i<m; i++)
for (int j=0; j<n; j++) {
if (rooms[i][j] == 0) {
queue.offer(new int[] {i,j});
}
}
int level = 1;
while (!queue.isEmpty()) {
int size = queue.size();
while (size > 0) {
int[] pt = queue.poll();
// rooms[pt[0]][pt[1]] = level;
for (int k = 0; k < 4; k++) {
int r = pt[0] + shift[k];
int c = pt[1] + shift[k+1];
if (r>=0 && r<m && c>=0 && c<n && rooms[r][c] == Integer.MAX_VALUE) {
rooms[r][c] = level; // set here in case point [1][1] would put [1][0] in the queue Twice
queue.offer(new int[] {r,c});
}
}
size--;
}
level++;
}
return;
}
}