# LeetCode – Walls and Gates (Java)

Java Solution 1 - DFS

```public void wallsAndGates(int[][] rooms) { if(rooms==null || rooms.length==0||rooms[0].length==0) return;   int m = rooms.length; int n = rooms[0].length;   boolean[][] visited = new boolean[m][n];   for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ if(rooms[i][j]==0){ fill(rooms, i-1, j, 0, visited); fill(rooms, i, j+1, 0, visited); fill(rooms, i+1, j, 0, visited); fill(rooms, i, j-1, 0, visited); visited = new boolean[m][n]; } } } }   public void fill(int[][] rooms, int i, int j, int start, boolean[][] visited){ int m=rooms.length; int n=rooms[0].length;   if(i<0||i>=m||j<0||j>=n||rooms[i][j]<=0||visited[i][j]){ return; }   rooms[i][j] = Math.min(rooms[i][j], start+1); visited[i][j]=true;   fill(rooms, i-1, j, start+1, visited); fill(rooms, i, j+1, start+1, visited); fill(rooms, i+1, j, start+1, visited); fill(rooms, i, j-1, start+1, visited);   visited[i][j]=false; }```

The DFS solution can be simplified as the following:

```public void wallsAndGates(int[][] rooms) { if(rooms==null || rooms.length==0||rooms[0].length==0) return;   int m = rooms.length; int n = rooms[0].length;   for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ if(rooms[i][j]==0){ fill(rooms, i, j, 0); } } } }   public void fill(int[][] rooms, int i, int j, int distance){ int m=rooms.length; int n=rooms[0].length;   if(i<0||i>=m||j<0||j>=n||rooms[i][j]<distance){ return; }   rooms[i][j] = distance;   fill(rooms, i-1, j, distance+1); fill(rooms, i, j+1, distance+1); fill(rooms, i+1, j, distance+1); fill(rooms, i, j-1, distance+1); }```

Java Solution 2 - BFS

```public void wallsAndGates(int[][] rooms) { if(rooms==null || rooms.length==0||rooms[0].length==0) return;   int m = rooms.length; int n = rooms[0].length;   LinkedList<Integer> queue = new LinkedList<Integer>();   for(int i=0; i<m; i++){ for(int j=0; j<n; j++){ if(rooms[i][j]==0){ queue.add(i*n+j); } } }   while(!queue.isEmpty()){ int head = queue.poll(); int x=head/n; int y=head%n;   if(x>0 && rooms[x-1][y]==Integer.MAX_VALUE){ rooms[x-1][y]=rooms[x][y]+1; queue.add((x-1)*n+y); }   if(x<m-1 && rooms[x+1][y]==Integer.MAX_VALUE){ rooms[x+1][y]=rooms[x][y]+1; queue.add((x+1)*n+y); }   if(y>0 && rooms[x][y-1]==Integer.MAX_VALUE){ rooms[x][y-1]=rooms[x][y]+1; queue.add(x*n+y-1); }   if(y<n-1 && rooms[x][y+1]==Integer.MAX_VALUE){ rooms[x][y+1]=rooms[x][y]+1; queue.add(x*n+y+1); } } }```
Category >> Algorithms
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• Yi-jhe Huang

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 2^31 – 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

• ambalaChambala

Can you post the actual question here.