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:
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。
1 public class Solution { 2 private int[][] dir = new int[][] { { -1, 0 }, { 1, 0 }, { 0, -1 }, { 0, 1 } }; 3 4 public int shortestDistance(int[][] grid) { 5 if (grid == null || grid.length == 0) { 6 return 0; 7 } 8 9 int rows = grid.length, cols = grid[0].length; 10 11 int[][] reach = new int[rows][cols]; 12 int[][] distance = new int[rows][cols]; 13 int numBuildings = 0; 14 15 // Find the minimum distance from all buildings 16 for (int i = 0; i < rows; i++) { 17 for (int j = 0; j < cols; j++) { 18 if (grid[i][j] == 1) { 19 shortestDistanceHelper(i, j, grid, reach, distance); 20 numBuildings++; 21 } 22 } 23 } 24 25 // step 2: check the min distance reachable by all buildings 26 int minDistance = Integer.MAX_VALUE; 27 for (int i = 0; i < rows; i++) { 28 for (int j = 0; j < cols; j++) { 29 if (grid[i][j] == 0 && reach[i][j] == numBuildings && distance[i][j] < minDistance) { 30 minDistance = distance[i][j]; 31 } 32 } 33 } 34 35 if (minDistance == Integer.MAX_VALUE) { 36 return -1; 37 } else { 38 return minDistance; 39 } 40 } 41 42 private void shortestDistanceHelper(int row, int col, int[][] grid, int[][] reach, int[][] distance) { 43 int rows = grid.length; 44 int cols = grid[0].length; 45 int d = 0; 46 boolean[][] visited = new boolean[rows][cols]; 47 Queue<int[]> queue = new LinkedList<>(); 48 queue.offer(new int[] { row, col }); 49 visited[row][col] = true; 50 while (!queue.isEmpty()) { 51 d++; 52 int size = queue.size(); 53 for (int j = 0; j < size; j++) { 54 int[] cord = queue.poll(); 55 for (int i = 0; i < 4; i++) { 56 int rr = dir[i][0] + cord[0]; 57 int cc = dir[i][1] + cord[1]; 58 if (isValid(rr, cc, grid, visited)) { 59 queue.offer(new int[] { rr, cc }); 60 visited[rr][cc] = true; 61 reach[rr][cc]++; 62 distance[rr][cc] += d; 63 } 64 } 65 } 66 } 67 } 68 69 private boolean isValid(int row, int col, int[][] grid, boolean[][] visited) { 70 int rows = grid.length; 71 int cols = grid[0].length; 72 73 if (row < 0 || row >= rows || col < 0 || col >= cols || visited[row][col] || grid[row][col] == 2) { 74 return false; 75 } 76 return true; 77 } 78 }
Shortest Distance from All Buildings
原文：https://www.cnblogs.com/beiyeqingteng/p/11261755.html