Leetcode: 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:

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

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。

public class Solution {
    public int minTotalDistance(int[][] grid) {
        List<Integer> xPos = new ArrayList<Integer>();
        List<Integer> yPos = new ArrayList<Integer>();
        for (int x=0; x<grid.length; x++) {
            for (int y=0; y<grid[0].length; y++) {
                if (grid[x][y] == 1) {
                    xPos.add(x);
                    yPos.add(y);
                }
            }
        }
        int minDis = 0;
        for (int x : xPos) {
            minDis += Math.abs(x - xPos.get(xPos.size()/2));
        }
        Collections.sort(yPos);
        for (int y : yPos) {
            minDis += Math.abs(y - yPos.get(yPos.size()/2));
        }
        return minDis;
    }
}