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.

横纵分离

复杂度

时间 O(NM) 空间 O(NM)

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。

代码

public class Solution {
    public int minTotalDistance(int[][] grid) {
        List<Integer> ipos = new ArrayList<Integer>();
        List<Integer> jpos = new ArrayList<Integer>();
        // 统计出有哪些横纵坐标
        for(int i = 0; i < grid.length; i++){
            for(int j = 0; j < grid[0].length; j++){
                if(grid[i][j] == 1){
                    ipos.add(i);
                    jpos.add(j);
                }
            }
        }
        int sum = 0;
        // 计算纵坐标到纵坐标中点的距离，这里不需要排序，因为之前统计时是按照i的顺序
        for(Integer pos : ipos){
            sum += Math.abs(pos - ipos.get(ipos.size() / 2));
        }
        // 计算横坐标到横坐标中点的距离，这里需要排序，因为统计不是按照j的顺序
        Collections.sort(jpos);
        for(Integer pos : jpos){
            sum += Math.abs(pos - jpos.get(jpos.size() / 2));
        }
        return sum;
    }
}