Shortest Distance from All Buildings

You want to build a house on an empty land which reaches all buildings in the shortest amount of distance. You can only move up, down, left and right. You are given a 2D grid of values 0, 1 or 2, where:

• Each 0 marks an empty land which you can pass by freely.
• Each 1 marks a building which you cannot pass through.
• Each 2 marks an obstacle which you cannot pass through.

For example, given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2):

1 - 0 - 2 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

The point (1,2) is an ideal empty land to build a house, as the total travel distance of 3+3+1=7 is minimal. So return 7.

Note:
There will be at least one building. If it is not possible to build such house according to the above rules, return -1.

思路：

从每个点是1的点出发，通过bfs找到这个点到每个0点最短距离，同时记录该0点被1点访问过的次数。这样我们遍历所有1点，对所有能够被访问到的1点，保存最短距离以及增加访问次数。最后把所有0点遍历一遍，看它是否被所有1点访问到，并且最短距离和最小。

其实这里也可以从0出发，做类似的事情。选1还是0看它们的个数。谁小就选谁。

 public class Solution {
     private int[][] dir = { { -,  }, { ,  }, { , - }, { ,  } };

     public int shortestDistance(int[][] grid) {
         if (grid == null || grid.length == ) {
             return ;
         }

         int rows = grid.length, cols = grid[].length, numBuildings = ;
         int[][] reach = new int[rows][cols], distance = new int[rows][cols];

         // Find the minimum distance from all buildings
         for (int i = ; i < rows; i++) {
             for (int j = ; j < cols; j++) {
                 if (grid[i][j] == ) {
                     shortestDistanceHelper(i, j, grid, reach, distance);
                     numBuildings++;
                 }
             }
         }

         // step 2: check the min distance reachable by all buildings
         int minDistance = Integer.MAX_VALUE;
         for (int i = ; i < rows; i++) {
             for (int j = ; j < cols; j++) {
                 if (grid[i][j] ==  && reach[i][j] == numBuildings && distance[i][j] < minDistance) {
                     minDistance = distance[i][j];
                 }
             }
         }
         return minDistance == Integer.MAX_VALUE ? - : minDistance;
     }

     private void shortestDistanceHelper(int row, int col, int[][] grid, int[][] reach, int[][] distance) {
         int rows = grid.length, cols = grid[].length, d = ;
         boolean[][] visited = new boolean[rows][cols];
         Queue<int[]> queue = new LinkedList<>();
         queue.offer(new int[] { row, col });
         visited[row][col] = true;
         while (!queue.isEmpty()) {
             d++;
             int size = queue.size();
             for (int j = ; j < size; j++) {
                 int[] cord = queue.poll();
                 for (int i = ; i < ; i++) {
                     int rr = dir[i][] + cord[];
                     int cc = dir[i][] + cord[];
                     if (isValid(rr, cc, grid, visited)) {
                         queue.offer(new int[] { rr, cc });
                         visited[rr][cc] = true;
                         reach[rr][cc]++;
                         distance[rr][cc] += d;
                     }
                 }
             }
         }
     }

     private boolean isValid(int row, int col, int[][] grid, boolean[][] visited) {
         int rows = grid.length, cols = grid[].length;
         if (row <  || row >= rows || col <  || col >= cols || visited[row][col] || grid[row][col] == ) {
             return false;
         }
         return true;
     }
 }