LeetCode 296. Best Meeting Point

原题链接在这里：https://leetcode.com/problems/best-meeting-point/

题目：

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|.

Example:

Input: 

1 - 0 - 0 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

Output: 6 

Explanation: Given three people living at (0,0), (0,4), and (2,2):
             The point (0,2) is an ideal meeting point, as the total travel distance 
             of 2+2+2=6 is minimal. So return 6.

题解：

When trying to minimize the manhattan distance, it is trying to minimize the absolute deviations of x and y.

And median minimize the absolute deviations.

We get all x and y when there is a building.

And calculate absolute deviations.

Time Complexity: O(m * n). m = grid.length. n = grid[0].length.

Space: O(m + n).

AC Java:

class Solution {
    public int minTotalDistance(int[][] grid) {
        if(grid == null || grid.length == 0 || grid[0].length == 0){
            return 0;
        }
        
        int m = grid.length;
        int n = grid[0].length;
        
        List<Integer> iIndexList = new ArrayList<>();
        for(int i = 0; i < m ; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    iIndexList.add(i);
                }
            }
        }
        
        List<Integer> jIndexList = new ArrayList<>();
        for(int j = 0; j < n; j++){
            for(int i = 0; i < m; i++){
                if(grid[i][j] == 1){
                    jIndexList.add(j);
                }
            }
        }
        
        int iDist = dist(iIndexList);
        int jDist = dist(jIndexList);
        return iDist + jDist; 
    }
    
    private int dist(List<Integer> list){
        int l = 0;
        int r = list.size() - 1;
        int res = 0;
        
        while(l < r){
            res += list.get(r--) - list.get(l++);
        }
        
        return res;
    }
}

类似Minimum Moves to Equal Array Elements II.