Shortest Distance from All Buildings
題意：
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.
解題思路：
難，很多細節要注意，記住map的xy座標跟我們一般二維陣列的座標不同，使用bfs來作。
dist[i][j] is the empty land (i, j) to all the buildings.
grid[i][j] is reused as the accessibility.
public class Solution {
public class Tuple {
private int y;
private int x;
private int dist;
public Tuple(int y, int x, int dist) {
this.y = y;
this.x = x;
this.dist = dist;
}
}
public int shortestDistance(int[][] grid) {
if (grid == null || grid.length == 0) {
return 0;
}
int m = grid.length;
int n = grid[0].length;
int[][] dist = new int[m][n];
List<Tuple> buildings = new ArrayList<>();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == 1) {
buildings.add(new Tuple(i, j, 0));
}
grid[i][j] = - grid[i][j];
}
}
for (int k = 0; k < buildings.size(); k++) {
bfs(buildings.get(k), k, dist, grid, m, n);
}
int ans = -1;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (grid[i][j] == buildings.size() && (ans < 0 || dist[i][j] < ans)) {
ans = dist[i][j];
}
}
}
return ans;
}
int[] idx = {0, 0, -1, 1};
int[] idy = {-1, 1, 0, 0};
public void bfs(Tuple root, int k, int[][] dist, int[][] grid, int m, int n) {
Queue<Tuple> q = new LinkedList<>();
q.offer(root);
while (!q.isEmpty()) {
Tuple cur = q.poll();
dist[cur.y][cur.x] += cur.dist;
for (int i = 0; i < 4; i++) {
int x = cur.x + idx[i];
int y = cur.y + idy[i];
if (y >= 0 && x >= 0 && y < m && x < n && grid[y][x] == k) {
grid[y][x] = k + 1;
q.offer(new Tuple(y, x, cur.dist + 1));
}
}
}
}
}