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.

Hint:

1. Try to solve it in one dimension first. How can this solution apply to the two dimension case?

思路：为了保证总长度最小，我们只要保证每条路径尽量不要重复就行了，比如1->2->3<-4这种一维的情况，如果起点是1，2和4，那2->3和1->2->3这两条路径就有重复了。为了尽量保证右边的点向左走，左边的点向右走，那我们就应该去这些点中间的点作为交点。由于是曼哈顿距离，我们可以分开计算横坐标和纵坐标，结果是一样的。所以我们算出各个横坐标到中点横坐标的距离，加上各个纵坐标到中点纵坐标的距离，就是结果了。

这里有个小的技巧，就是计算到中间值的距离，只需计算后面的点减去前面的点，然后i++,j--就行了，因为是计算每个点移动到中间的距离。

public class Solution {    public int minTotalDistance(int[][] grid) {        List<Integer> rowlist = new ArrayList<Integer>();        List<Integer> collist = new ArrayList<Integer>();        for(int i=0; i<grid.length; i++) {            for(int j=0; j<grid[0].length; j++){                if(grid[i][j] == 1){                    rowlist.add(i);                    collist.add(j);                }            }        }                return calMedian(rowlist) + calMedian(collist);    }        public int calMedian(List<Integer> list) {        Collections.sort(list);        int i=0; int j=list.size()-1;        int res = 0;        while(i<j){            res += list.get(j) - list.get(i);            i++;            j--;        }        return res;    }}