286. 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 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.

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

Understand the problem:
It is very classic backtracking problem. We can start from each gate (0 point), and searching for its neighbors. We can either use DFS or BFS solution.

public class Solution {
    public void wallsAndGates(int[][] rooms) {
        if (rooms == null || rooms.length == 0) return;
        int m = rooms.length, n = rooms[0].length;
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (rooms[i][j] == 0) {
                    helper(i, j, 0, rooms);
                }
            }
        }
    }
     
    private void helper(int row, int col, int distance, int[][] rooms) {
        int rows = rooms.length, cols = rooms[0].length;
        if (row < 0 || row >= rows || col < 0 || col >= cols || rooms[row][col] == -1) return;
        if (distance > rooms[row][col]) return;
        if (distance < rooms[row][col]) {
            rooms[row][col] = distance;
        }
        helper(row - 1, col, distance + 1, rooms);
        helper(row + 1, col, distance + 1, rooms);
        helper(row, col - 1, distance + 1, rooms);
        helper(row, col + 1, distance + 1, rooms);
    }
}

 

BFS

对于bfs，因为我们是把所有的0点在最开始的时候加入到了queue里面，所以，当其中一个0点访问到一个空点的时候，那么我们一定可以说那是最短距离。所以，时间复杂度上，bfs比dfs好很多。

public class Solution {
    public void wallsAndGates(int[][] rooms) {
        Queue<int[]> queue = new LinkedList<int[]>();
        int rows = rooms.length;
        if (rows == 0) {
            return;
        }
        int cols = rooms[0].length;
        
        // 找出所有BFS的起始点
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (rooms[i][j] == 0) {
                    queue.offer(new int[]{i, j});
                }
            }
        }
        
        // 定义下一步的位置
        int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        
        // 开始BFS
        while (!queue.isEmpty()) {
            int[] top = queue.poll();
            for (int k = 0; k < dirs.length; k++) {
                int x = top[0] + dirs[k][0];
                int y = top[1] + dirs[k][1];
                if (x >= 0 && x < rows && y >= 0 && y < cols && rooms[x][y] == Integer.MAX_VALUE) {
                    rooms[x][y] = rooms[top[0]][top[1]] + 1;
                    queue.add(new int[]{x, y});
                }
            }
        }
    }
}

From:

http://buttercola.blogspot.com/2015/09/leetcode-walls-and-gates.html

https://segmentfault.com/a/1190000004184488