317. 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.

一刷
思路： 通过BFS, 以building为起点计算出每个empty space到所有building的距离，然后从其中取出最小的即为满足要求的点。

class Solution {
    class Tuple{
        int x;
        int y;
        int dist;
        Tuple(int x, int y, int dist){
            this.x = x;
            this.y = y;
            this.dist = dist;
        }
    }
    
    int[][] dirs = {{1, 0},{-1, 0},{0, 1},{0, -1}};
    
    public int shortestDistance(int[][] grid) {
        int m = grid.length, n = grid[0].length;
        List building = new ArrayList<>();
        
        for(int i=0; i queue = new ArrayDeque<>();
        queue.add(root);
        while(!queue.isEmpty()){
            Tuple cur = queue.poll();
            dist[cur.x][cur.y] += cur.dist;
            for(int[] dir : dirs){
                int x = cur.x + dir[0];
                int y = cur.y + dir[0];
                if(x>=0 && y>=0 && x