-
Notifications
You must be signed in to change notification settings - Fork 0
Walls and Gates
Tim_Gao edited this page Sep 17, 2016
·
2 revisions
-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.length == 0 || rooms[0].length == 0){
return;
}
LinkedList<int[]> q1 = new LinkedList<>(), q2 = new LinkedList<>();
for (int i=0; i<rooms.length; ++i){
for (int j=0; j<rooms[0].length; ++j){
if(rooms[i][j] == 0){
q1.add(new int[]{i, j});
}
}
}
int dis = 1;
while (!q1.isEmpty()){
q2 = q1;
q1 = new LinkedList<>();
while (!q2.isEmpty()){
int[] crt = q2.getFirst();
q2.removeFirst();
int x = crt[0], y = crt[1];
if(x-1>=0 && rooms[x-1][y] == Integer.MAX_VALUE){
rooms[x-1][y] = dis;
q1.add(new int[]{x-1, y});
}
if(x+1 < rooms.length && rooms[x+1][y] == Integer.MAX_VALUE){
rooms[x+1][y] = dis;
q1.add(new int[]{x+1, y});
}
if (y-1>=0 && rooms[x][y-1] == Integer.MAX_VALUE){
rooms[x][y-1] = dis;
q1.add(new int[]{x, y-1});
}
if (y+1<rooms[0].length && rooms[x][y+1] == Integer.MAX_VALUE){
rooms[x][y+1] = dis;
q1.add(new int[]{x, y+1});
}
}
++dis;
}
}
}