Shortest Distance from All Buildings

最新推荐文章于 2025-09-10 21:15:00 发布

转载最新推荐文章于 2025-09-10 21:15:00 发布 · 54 阅读

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原文链接：http://www.cnblogs.com/shuashuashua/p/5642236.html

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#开发工具

本文介绍了一种算法，该算法用于解决在给定网格中找到空地以便最小化到所有建筑物的距离之和的问题。通过从每个建筑物出发进行层级遍历，计算每个空地到所有建筑物的总距离，并最终找出符合条件的最小距离。

 1 public class Solution {
 2     public int shortestDistance(int[][] grid) {
 3         if (grid.length == 0 || grid[0].length == 0) {
 4             return -1;
 5         }
 6         
 7         int[] shift = new int[]{0, 1, 0, -1, 0};
 8         int[][] distance = new int[grid.length][grid[0].length];
 9         int[][] reach = new int[grid.length][grid[0].length];
10         int numBuildings = 0;
11         for (int i = 0; i < grid.length; i++) {
12             for (int j = 0; j < grid[0].length; j++) {
13                 if (grid[i][j] == 1) {
14                     int level = 1;
15                     numBuildings++;
16                     boolean[][] visited = new boolean[grid.length][grid[0].length];
17                     Queue<int[]> queue = new LinkedList<>();
18                     queue.offer(new int[]{i, j});
19                     while (!queue.isEmpty()) {
20                         int size = queue.size();
21                         for (int k = 0; k < size; k++) {
22                             int[] current = queue.poll();
23                             for (int l = 0; l < 4; l++) {
24                                 int x = current[0] + shift[l];
25                                 int y = current[1] + shift[l + 1];
26                                 if (x >= 0 && x < grid.length && y >= 0 && y < grid[0].length && !visited[x][y] && grid[x][y] == 0) {
27                                     visited[x][y] = true;
28                                     distance[x][y] += level;
29                                     reach[x][y]++;
30                                     queue.offer(new int[]{x, y});
31                                 }
32                             }
33                         }
34                         level++;
35                     }
36                 }
37             }
38         }
39         
40         int result = Integer.MAX_VALUE;
41         for (int i = 0; i < grid.length; i++) {
42             for (int j = 0; j < grid[0].length; j++) {
43                 if (grid[i][j] == 0 && reach[i][j] == numBuildings) {
44                     result = Math.min(result, distance[i][j]);
45                 }
46             }
47         }
48         return result == Integer.MAX_VALUE ? -1 : result;
49     }
50 }