原题链接在这里:https://leetcode.com/problems/shortest-distance-from-all-buildings/

题目:

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.

Example:

Input: [[1,0,2,0,1],[0,0,0,0,0],[0,0,1,0,0]]

1 - 0 - 2 - 0 - 1
| | | | |
0 - 0 - 0 - 0 - 0
| | | | |
0 - 0 - 1 - 0 - 0 Output: 7 Explanation: Given three buildings at (0,0), (0,4), (2,2), and an obstacle at (0,2),
t
he 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.

题解:

从每一座building开始做BFS, 更新每个空地达到building的距离总和 以及 每个空地能到达building的个数.

第二次扫描grid, 若是空地并且它能到达的building数目是总共的building数目,就更新min距离.

Note: For the second iteration, check 2 conditions. grid[i][j] < 0 && reachCount[i][j] = totalCount.

Time Complexity: O(m^2 * n^2), 每次BFS用O(mn), 一共做了m*n次BFS.

Space: O(m*n)

AC Java:

 class Solution {
public int shortestDistance(int[][] grid) {
if(grid == null || grid.length == 0 || grid[0].length == 0){
return 0;
} int m = grid.length;
int n = grid[0].length; //记录每个点能够到达building的个数
int [][] reachCount = new int[m][n];
int totalCount = 0; for(int i = 0; i<m; i++){
for(int j = 0; j<n; j++){
if(grid[i][j] == 1){
//遇到building, 从这个building开始做bfs
totalCount++;
bfs(grid, i, j, reachCount);
}
}
} int res = Integer.MAX_VALUE;
for(int i = 0; i<m; i++){
for(int j = 0; j<n; j++){
if(grid[i][j] < 0 && reachCount[i][j] == totalCount){
res = Math.min(res, -grid[i][j]);
}
}
} return res == Integer.MAX_VALUE ? -1 : res;
} int [][] dirs = new int[][]{{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
private void bfs(int [][] grid, int i, int j, int [][] reachCount){
int level = 0;
int m = grid.length;
int n = grid[0].length;
LinkedList<int []> que = new LinkedList<>();
boolean[][] visited = new boolean[m][n];
que.add(new int[]{i, j});
visited[i][j] = true; while(!que.isEmpty()){
int size = que.size();
while(size-- > 0){
int [] cur = que.poll();
grid[cur[0]][cur[1]] -= level;
reachCount[cur[0]][cur[1]]++; for(int [] dir : dirs){
int x = cur[0] + dir[0];
int y = cur[1] + dir[1];
if(x < 0 || x >=m || y < 0 || y >=n || visited[x][y] || grid[x][y] > 0){
continue;
} que.add(new int[]{x, y});
visited[x][y] = true;
}
} level++;
}
}
}

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