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:

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

Understand the problem:
http://segmentfault.com/a/1190000003894693

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。

Code (Java):

public class Solution {
    public int minTotalDistance(int[][] grid) {
        if (grid == null || grid.length == 0) {
            return 0;
        }
        
        int m = grid.length;
        int n = grid[0].length;
        
        List<Integer> rowIndex = new ArrayList<>();
        List<Integer> colIndex = new ArrayList<>();
        
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    rowIndex.add(i);
                    colIndex.add(j);
                }
            }
        }
        
        int sum = 0;
        int mid = rowIndex.get(rowIndex.size() / 2);
        for (int x : rowIndex) {
            sum += Math.abs(x - mid);
        }
        
        Collections.sort(colIndex);
        mid = colIndex.get(colIndex.size() / 2);
        
        for (int y : colIndex) {
            sum += Math.abs(y - mid);
        }
        
        return sum;
    }
}