296. Best Meeting Point

A group of two or more people wants to meet and minimize the total travel distance. You are given a 2D grid of values 0 or 1, where each 1 marks the home of someone in the group. The distance is calculated using Manhattan Distance, where distance(p1, p2) = |p2.x - p1.x| + |p2.y - p1.y|.

For example, given three people living at (0,0)(0,4), and (2,2):

1 - 0 - 0 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

The point (0,2) is an ideal meeting point, as the total travel distance of 2+2+2=6 is minimal. So return 6.

 

本题可以归结为数学问题,让我们求最佳的开会地点,A————P————B,从一维上面来考虑,假如P为最佳的点,那么P一定是在AB的中间,如果在左面或者右面,则距离就大于AB的距离了。四个点的情况如下:C———A——P——B——D,P也一定位于AB之间,而三个点的情况C——A——B,P一定位于刚好A点的距离,此时距离最小。扩展二维距离也是类似的,实现的时候需要注意,一定要把点分别按照row从小到大和col从小到大排序。

代码如下:

 1 public class Solution {
 2     public int minTotalDistance(int[][] grid) {
 3         int m = grid.length;
 4         int n = grid[0].length;
 5         List<Integer> row = new ArrayList<>();
 6         List<Integer> col = new ArrayList<>();
 7         for(int i=0;i<m;i++){
 8             for(int j=0;j<n;j++){
 9                 if(grid[i][j]==1){
10                     row.add(i);
11                     col.add(j);
12                 }
13             }
14         }
15         return getMin(row)+getMin(col);
16     }
17     public int getMin(List<Integer> list){
18         int i=0;
19         int j=list.size()-1;
20         Collections.sort(list);
21         int res = 0;
22         while(i<j){
23             res+=list.get(j--)-list.get(i++);
24         }
25         return res;
26     }
27 }
28 // the run time of the O(mnlogn).the space complexity could be O(m+n);

 

posted @ 2017-05-25 00:55  CodesKiller  阅读(188)  评论(0编辑  收藏  举报