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.
分析
这道理类似于Walls ang Gates, 解决方法也应该是从building出发进行BFS,不过这里不同的是这里需要返回最小距离和,所以我们应该一个一个的对building的点BFS,用一个二维矩阵存每个点到所有building的距离和,每次BFS,都更新相应的距离和。最后遍历那个距离和矩阵,找出最小值即可。
需要注意的是,这道题还有个条件就是empty room 必须 reach all buildings,所以我们可以用另外一个矩阵存对应empty room到building的个数,如果最终个数不等于总的building数,对应点存的距离和无效。
复杂度
time: O(kMN), space: O(MN), k表示building数量
代码
public class Solution {
public int shortestDistance(int[][] grid) {
int rows = grid.length;
if (rows == 0) {
return -1;
}
int cols = grid[0].length;
// 记录到各个building距离和
int[][] dist = new int[rows][cols];
// 记录到能到达的building的数量
int[][] nums = new int[rows][cols];
int buildingNum = 0;
// 从每个building开始BFS
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (grid[i][j] == 1) {
buildingNum++;
bfs(grid, i, j, dist, nums);
}
}
}
int min = Integer.MAX_VALUE;
for (int i = 0; i < rows; i++) {
for (int j = 0; j < cols; j++) {
if (grid[i][j] == 0 && dist[i][j] != 0 && nums[i][j] == buildingNum)
min = Math.min(min, dist[i][j]);
}
}
if (min < Integer.MAX_VALUE)
return min;
return -1;
}
public void bfs(int[][] grid, int row, int col, int[][] dist, int[][] nums) {
int rows = grid.length;
int cols = grid[0].length;
Queue<int[]> q = new LinkedList<>();
q.add(new int[]{row, col});
int[][] dirs = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
// 记录访问过的点
boolean[][] visited = new boolean[rows][cols];
int level = 0;
while (!q.isEmpty()) {
level++;
int size = q.size();
for (int i = 0; i < size; i++) {
int[] coords = q.remove();
for (int k = 0; k < dirs.length; k++) {
int x = coords[0] + dirs[k][0];
int y = coords[1] + dirs[k][1];
if (x >= 0 && x < rows && y >= 0 && y < cols && !visited[x][y] && grid[x][y] == 0) {
visited[x][y] = true;
dist[x][y] += level;
nums[x][y]++;
q.add(new int[]{x, y});
}
}
}
}
}
}
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