# 296 Best Meeting Point

## 296. [Best Meeting Point](https://leetcode.com/problems/best-meeting-point/description/)

## 1. Question

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](http://en.wikipedia.org/wiki/Taxicab_geometry), 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

## 2. Implementation

**(1) Sort + Two Pointers**

思路: 和#462的思路一样, 相遇的点在哪不影响最后的距离

```java
class Solution {
    public int minTotalDistance(int[][] grid) {
        List<Integer> rows = new ArrayList<>();
        List<Integer> cols = new ArrayList<>();

        int m = grid.length, n = grid[0].length;

        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    rows.add(i);
                    cols.add(j);
                }
            }
        }
        return getDistance(rows, false) + getDistance(cols, true);
    }

    public int getDistance(List<Integer> list, boolean isCol) {
        // rows这个list本身已经sorted，所以只对cols排序
        if (isCol) {
           Collections.sort(list); 
        }

        int start = 0, end = list.size() - 1;
        int dist = 0;
        while (start < end) {
            dist += list.get(end) - list.get(start);
            ++start;
            --end;
        }
        return dist;
    }
}
```

## 3. Time & Space Complexity

**Sort + Two Pointers:** 时间复杂度O(mn + mlogm + nlogn), 空间复杂度O(Max(m,n))