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 == ) return;
         int m = rooms.length, n = rooms[].length;

         for (int i = ; i < m; i++) {
             for (int j = ; j < n; j++) {
                 if (rooms[i][j] == ) {
                     helper(i, j, , rooms);
                 }
             }
         }
     }

     private void helper(int row, int col, int distance, int[][] rooms) {
         int rows = rooms.length, cols = rooms[].length;
         if (row <  || row >= rows || col <  || col >= cols || rooms[row][col] == -) return;
         if (distance > rooms[row][col]) return;
         if (distance < rooms[row][col]) {
             rooms[row][col] = distance;
         }
         helper(row - , col, distance + , rooms);
         helper(row + , col, distance + , rooms);
         helper(row, col - , distance + , rooms);
         helper(row, col + , distance + , 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 == ) {
             return;
         }
         int cols = rooms[].length;

         // 找出所有BFS的起始点
         for (int i = ; i < rows; i++) {
             for (int j = ; j < cols; j++) {
                 if (rooms[i][j] == ) {
                     queue.offer(new int[]{i, j});
                 }
             }
         }

         // 定义下一步的位置
         int[][] dirs = {{-, }, {, }, {, -}, {, }};

         // 开始BFS
         while (!queue.isEmpty()) {
             int[] top = queue.poll();
             for (int k = ; k < dirs.length; k++) {
                 int x = top[] + dirs[k][];
                 int y = top[] + dirs[k][];
                 if (x >=  && x < rows && y >=  && y < cols && rooms[x][y] == Integer.MAX_VALUE) {
                     rooms[x][y] = rooms[top[]][top[]] + ;
                     queue.add(new int[]{x, y});
                 }
             }
         }
     }
 }

From:

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

https://segmentfault.com/a/1190000004184488