286. Walls and Gates
You are given an m x n
grid rooms
initialized with these three possible values.
-1
A wall or an obstacle.0
A gate.INF
Infinity means an empty room. We use the value231 - 1 = 2147483647
to representINF
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
.
Example 1:
Input: rooms = [[2147483647,-1,0,2147483647],[2147483647,2147483647,2147483647,-1],[2147483647,-1,2147483647,-1],[0,-1,2147483647,2147483647]]
Output: [[3,-1,0,1],[2,2,1,-1],[1,-1,2,-1],[0,-1,3,4]]
Example 2:
Constraints:
m == rooms.length
n == rooms[i].length
1 <= m, n <= 250
rooms[i][j]
is-1
,0
, or231 - 1
.
Solution:
class Solution {
static class Node{
int row;
int col;
int distance;
Node(int row, int col, int distance){
this.row = row;
this.col = col;
this.distance = distance;
}
}
public void wallsAndGates(int[][] rooms) {
Queue<Node> 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 Node(i, j, 0));
}
}
}
while(!queue.isEmpty()){
Node cur = queue.poll();
helper(cur.row - 1, cur.col, cur.distance, rooms, queue);
helper(cur.row, cur.col - 1, cur.distance, rooms, queue);
helper(cur.row + 1, cur.col, cur.distance, rooms, queue);
helper(cur.row, cur.col + 1, cur.distance, rooms, queue);
}
}
private void helper(int row, int col, int distance, int[][] rooms, Queue<Node> queue){
if (row < 0 || row >= rooms.length || col < 0 || col >= rooms[0].length || rooms[row][col] != 2147483647){
return;
}
if (rooms[row][col] > distance + 1){
rooms[row][col] = distance + 1;
queue.add(new Node(row, col, distance + 1));
}
}
}
// TC: O(n^2)
// SC: O(n^2)