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 value231 - 1 = 2147483647to representINFas 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:
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
典型的BFS,注意的是需要从所有0的位置同时开始走
O(m * n), m n 为2维数组的长宽。因为一个INF的点最多只被走一次
class Solution {
public void wallsAndGates(int[][] rooms) {
Queue<Pair> que = getQueue(rooms);
fillRooms(rooms, que);
}
private void fillRooms(int[][] rooms, Queue<Pair> que) {
int[][] coord = {{1,0}, {-1,0}, {0,1}, {0,-1}};
while (!que.isEmpty()) {
Pair top = que.poll();
for (int[] arr : coord) {
int r = top.row + arr[0];
int c = top.col + arr[1];
if (r >= 0 && r < rooms.length && c >= 0 && c < rooms[0].length
&& rooms[r][c] == 2147483647) {
rooms[r][c] = rooms[top.row][top.col] + 1;
que.add(new Pair(r, c));
}
}
}
}
private Queue<Pair> getQueue(int[][] rooms) {
Queue<Pair> que = new LinkedList<>();
for (int i = 0; i < rooms.length; i++) {
for (int j = 0; j < rooms[0].length; j++) {
if (rooms[i][j] == 0) {
que.add(new Pair(i, j));
}
}
}
return que;
}
}
class Pair{
public int row;
public int col;
public Pair(int row, int col) {
this.row = row;
this.col = col;
}
}
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