Shortest Distance from All Buildings.java

H
1533539906
tags: BFS

给Walls and Gates很像, 不同的是, 这道题要选一个 coordinate, having shortest sum distance to all buildings (marked as 1).

#### BFS
- BFS 可以 mark shortest distance from bulding -> any possible spot.
- Try each building (marked as 1) -> BFS cover all 0. 
- time: O(n^2) * # of building; use new visited[][] to mark visited for each building.
- O(n^2) find smallest point/aggregation value.
- 注意, 这道题我们update grid[][] sum up with shortest path value from building.
- 最后找个min value 就好了, 甚至不用return coordinate.
- 分析过, 还没有写.

```

/*
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),
             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.

 */
```