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：

public class Solution {
    public int minTotalDistance(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;
        List<Integer> row = new ArrayList<>();
        List<Integer> col = new ArrayList<>();
        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    row.add(i);
                    col.add(j);
                }
            }
        }
        return getMin(row)+getMin(col);
    }
    public int getMin(List<Integer> list){
        int i=0;
        int j=list.size()-1;
        Collections.sort(list);
        int res = 0;
        while(i<j){
            res+=list.get(j--)-list.get(i++);
        }
        return res;
    }
}
// the run time of the O(mnlogn).the space complexity could be O(m+n);