LeetCode – Walls and Gates (Java)

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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);
        }
    }
}

2 thoughts on “LeetCode – Walls and Gates (Java)”

  1. 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

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