Walls and Gates

You are given a m x n 2D grid initialized with these three possible values.

1. -1 – A wall or an obstacle.

2. 0 – A gate.

3. 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

广度优先搜索

复杂度

时间 O(NM) 空间 O(N)

思路

实际上就是找每个房间到最近的门的距离，我们从每个门开始，广度优先搜索并记录层数就行了。如果某个房间之前被标记过距离，那就选择这个距离和当前距离中较小的那个。这题要注意剪枝，如果下一步是门或者下一步是墙或者下一步已经访问过了，就不要加入队列中。否则会超时。

代码

public class Solution {
    public void wallsAndGates(int[][] rooms) {
        if(rooms.length == 0) return;
        for(int i = 0; i < rooms.length; i++){
            for(int j = 0; j < rooms[0].length; j++){
                // 如果遇到一个门，从门开始广度优先搜索，标记连通的节点到自己的距离
                if(rooms[i][j] == 0) bfs(rooms, i, j);
            }
        }
    }
    
    public void bfs(int[][] rooms, int i, int j){
        Queue<Integer> queue = new LinkedList<Integer>();
        queue.offer(i * rooms[0].length + j);
        int dist = 0;
        // 用一个集合记录已经访问过的点
        Set<Integer> visited = new HashSet<Integer>();
        visited.add(i * rooms[0].length + j);
        while(!queue.isEmpty()){
            int size = queue.size();
            // 记录深度的搜索
            for(int k = 0; k < size; k++){
                Integer curr = queue.poll();
                int row = curr / rooms[0].length;
                int col = curr % rooms[0].length;
                // 选取之前标记的值和当前的距离的较小值
                rooms[row][col] = Math.min(rooms[row][col], dist);
                int up = (row - 1) * rooms[0].length + col;
                int down = (row + 1) * rooms[0].length + col;
                int left = row * rooms[0].length + col - 1;
                int right = row * rooms[0].length + col + 1;
                if(row > 0 && rooms[row - 1][col] > 0 && !visited.contains(up)){
                    queue.offer(up);
                    visited.add(up);
                }
                if(col > 0 && rooms[row][col - 1] > 0 && !visited.contains(left)){
                    queue.offer(left);
                    visited.add(left);
                }
                if(row < rooms.length - 1 && rooms[row + 1][col] > 0 && !visited.contains(down)){
                    queue.offer(down);
                    visited.add(down);
                }
                if(col < rooms[0].length - 1 && rooms[row][col + 1] > 0 && !visited.contains(right)){
                    queue.offer(right);
                    visited.add(right);
                }
            }
            dist++;
        }
    }
}