Build Post Office II

Given a 2D grid, each cell is either a wall 2, an house 1 or empty 0 (the number zero, one, two), find a place to build a post office so that the sum of the distance from the post office to all the houses is smallest.

Return the smallest sum of distance. Return -1 if it is not possible.

class Coordinate {    int x;    int y;    public Coordinate(int x, int y) {        this.x = x;        this.y = y;    }}public class Solution {    /**     * @param grid a 2D grid     * @return an integer     */    public int EMPTY = 0;    public int HOUSE = 1;    public int WALL = 2;    public int n;    public int m;    public int[][] grid;    //四个方向    public int[] directionX = {0, 0, 1, -1};    public int[] directionY = {1, -1, 0, 0};    //初始化n, m, grid    private void setGrid(int[][] grid) {        this.n = grid.length;        this.m = grid[0].length;        this.grid = grid;    }    //获取1，0的坐标序列    private ArrayList<Coordinate> getCoordinates(int type) {        ArrayList<Coordinate> coordinates = new ArrayList<>();        for (int i = 0; i < n; i++) {            for (int j = 0; j < m; j++) {                if (grid[i][j] == type) {                    coordinates.add(new Coordinate(i, j));                }            }        }        return coordinates;    }    //判断坐标是否valid    private boolean inBound(int x, int y, int[][] grid) {        int n = grid.length;        int m = grid[0].length;        return x >= 0 && x < n && y >= 0 && y < m && grid[x][y] == 0;    }    //从房子出发，累加房子到每个空白的距离    public int shortestDistance(int[][] grid) {        if (grid == null || grid.length == 0 || grid[0].length == 0) {            return -1;        }        setGrid(grid);        ArrayList<Coordinate> house = getCoordinates(HOUSE);        //累加数组        int[][] distanceSum = new int[n][m];        //标记房子来过的记录，如果最后记录！=房子的数目，则说明有些空白不可达        int[][] visited = new int[n][m];        for (Coordinate h : house) {            bfsHelper(h, distanceSum, visited);        }        int shortest = Integer.MAX_VALUE;        ArrayList<Coordinate> empty = getCoordinates(EMPTY);        for (Coordinate e : empty) {            if (visited[e.x][e.y] != house.size()) {                continue;            }            shortest = Math.min(shortest, distanceSum[e.x][e.y]);        }        if (shortest == Integer.MAX_VALUE) {            return -1;        }        return shortest;    }    private void bfsHelper(Coordinate target, int[][] distanceSum, int[][] visited) {        Queue<Coordinate> queue = new LinkedList<>();        queue.offer(new Coordinate(target.x, target.y));        boolean[][] hash = new boolean[n][m];        hash[target.x][target.y] = true;        int step = 0;        while (!queue.isEmpty()) {            step++;            int size = queue.size();            for (int i = 0; i < size; i++) {                Coordinate coor = queue.poll();                for (int j = 0; j < 4; j++) {                    Coordinate adj = new Coordinate(                        coor.x + directionX[j],                        coor.y + directionY[j]                    );                    if (!inBound(adj.x, adj.y, grid)) {                        continue;                    }                    if (hash[adj.x][adj.y]) {                        continue;                    }                    queue.offer(adj);                    distanceSum[adj.x][adj.y] += step;                    visited[adj.x][adj.y]++;                    hash[adj.x][adj.y] = true;                }            }        }    }}