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
Solution1:BFS
Time Complexity: O(mn) Space Complexity: O(mn)
Solution1.b:BFS level(dist)
Time Complexity: O(mn) Space Complexity: O(mn)
Solution2:DFS染色
Time Complexity: O(mn) Space Complexity: O(mn)
Solution1 Code:
class Solution {
public void wallsAndGates(int[][] rooms) {
if (rooms.length == 0 || rooms[0].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.offer(new int[]{i, j});
}
}
while (!queue.isEmpty()) {
int[] top = queue.poll();
int row = top[0], col = top[1];
if (row > 0 && rooms[row - 1][col] == Integer.MAX_VALUE) {
rooms[row - 1][col] = rooms[row][col] + 1;
queue.offer(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.offer(new int[]{row + 1, col});
}
if (col > 0 && rooms[row][col - 1] == Integer.MAX_VALUE) {
rooms[row][col - 1] = rooms[row][col] + 1;
queue.offer(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.offer(new int[]{row, col + 1});
}
}
}
}
Solution1.b Code:
class Solution {
public void wallsAndGates(int[][] rooms) {
if(rooms == null || rooms.length == 0 || rooms[0].length == 0) return;
Queue<int[]> queue = new LinkedList<>();
for(int row = 0; row < rooms.length; row++) {
for(int col = 0; col < rooms[0].length; col++) {
if(rooms[row][col] == 0) {
queue.offer(new int[]{row, col});
}
}
}
int dist = 0;
while(!queue.isEmpty()) {
dist++;
int queue_size = queue.size();
for(int l = 0; l < queue_size; l++) {
int[] coord = queue.poll();
int[] step_row = new int[] {0, 0, -1, 1};
int[] step_col = new int[] {-1, 1, 0, 0};
for(int i = 0; i < 4; i++) {
int[] new_coord = new int[]{coord[0] + step_row[i], coord[1] + step_col[i]};
if(new_coord[0] < 0 || new_coord[0] > rooms.length - 1 || new_coord[1] < 0
|| new_coord[1] > rooms[0].length - 1) {
continue;
}
if(rooms[new_coord[0]][new_coord[1]] == Integer.MAX_VALUE) {
rooms[new_coord[0]][new_coord[1]] = dist;
queue.offer(new_coord);
}
}
}
}
}
}
Solution2 Code:
class Solution {
public void wallsAndGates(int[][] rooms) {
if(rooms.length==0||rooms[0].length==0)
return;
int n = rooms.length, m = rooms[0].length;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(rooms[i][j]==0)
DFS(rooms, i, j, 0);
}
}
}
public void DFS(int[][] rooms, int i, int j, int step){
if(i<0||j<0||i>=rooms.length||j>=rooms[0].length||rooms[i][j]==-1||rooms[i][j]<step)
return;
rooms[i][j] = step;
DFS(rooms, i+1, j, step+1);
DFS(rooms, i-1, j, step+1);
DFS(rooms, i, j+1, step+1);
DFS(rooms, i, j-1, step+1);
}
}