317 Shortest Distance from All Buildings
1. Question
You want to build a house on anemptyland 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.
2. Implementation
(1) BFS
class Solution {
public int shortestDistance(int[][] grid) {
if (grid == null || grid.length == 0) {
return -1;
}
int m = grid.length;
int n = grid[0].length;
int[][] directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
int[][] distance = new int[m][n];
int[][] reach = new int[m][n];
int buildings = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
++buildings;
getDistanceByBFS(i, j, distance, reach, directions, grid);
}
}
}
int res = Integer.MAX_VALUE;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 0 && reach[i][j] == buildings) {
res = Math.min(res, distance[i][j]);
}
}
}
return res == Integer.MAX_VALUE ? -1 : res;
}
public void getDistanceByBFS(int row, int col, int[][] distance, int[][] reach, int[][] directions, int[][] grid) {
int m = grid.length;
int n = grid[0].length;
boolean[][] visited = new boolean[m][n];
Queue<int[]> queue = new LinkedList<>();
queue.add(new int[] {row, col});
int dist = 0;
while (!queue.isEmpty()) {
++dist;
int size = queue.size();
for (int i = 0; i < size; i++) {
int[] curCell = queue.remove();
for (int[] direction : directions) {
int nextRow = curCell[0] + direction[0];
int nextCol = curCell[1] + direction[1];
if (isValid(nextRow, nextCol, visited, grid)) {
distance[nextRow][nextCol] += dist;
reach[nextRow][nextCol] += 1;
queue.add(new int[] {nextRow, nextCol});
visited[nextRow][nextCol] = true;
}
}
}
}
}
public boolean isValid(int row, int col, boolean[][] visited, int[][] grid) {
return row >= 0 && row < grid.length && col >= 0 && col < grid[0].length && !visited[row][col] && grid[row][col] == 0;
}
}
3. Time & Space Complexity
BFS: 时间复杂度O(M^2N^2),BFS的时间复杂度是O(mn), 总的时间复杂度是 O(# of buildings ) * O(mn), 最坏的情况是每个cell都是building, 总的时间复杂度为O(mn) * O(mn) => O(m^2n^2), 空间复杂度O(mn)
Last updated
Was this helpful?