296 Best Meeting Point

296. Best Meeting Point

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, 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

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))