Best Meeting Point 解答

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.

Hint:

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

Solution 1 -- Sort

根据曼哈顿距离的表达式，我们可以把这个问题转化为求每一维的最短距离。

总距离 ＝ x上最短总距离 ＋ y上最短总距离

并且，我们观察到，对于一个序列，如 [0,0,1,0,1,0,1,1,1,0,0,1]

当邮局选在median的位置时，距离和是最小的。

因此，这里我们遍历数组，得到home的x和y坐标值，然后分别对这两个list排序。再用双指针得到总的最短距离和。

Time complexity O(mn log(mn))

 public class Solution {
     public int minTotalDistance(int[][] grid) {
         if (grid == null || grid.length < 1) {
             return 0;
         }
         int m = grid.length, n = grid[0].length;
         List<Integer> xList = new ArrayList<Integer>();
         List<Integer> yList = new ArrayList<Integer>();
         for (int i = 0; i < m; i++) {
             for (int j = 0; j < n; j++) {
                 if (grid[i][j] == 1) {
                     xList.add(i);
                     yList.add(j);
                 }
             }
         }
         return calcTotalDistance(xList) + calcTotalDistance(yList);
     }

     private int calcTotalDistance(List<Integer> list) {
         if (list == null || list.size() < 2) {
             return 0;
         }
         Collections.sort(list);
         int len = list.size();
         int i = 0, j = len - 1;
         int result = 0;
         while (i < j) {
             int left = list.get(i);
             int right = list.get(j);
             result += (right - left);
             i++;
             j--;
         }
         return result;
     }
 }

Solution 2 -- Without sorting

Discussion里有人给出了不用排序的方法。时间复杂度降低为O(mn)

因为双层循环遍历数组本身就是有次序性的。

1. 外层循环为遍历行，内层循环为遍历该行的每一个元素。

这样得到的是排序好的x的list

2. 外层循环为遍历列，内层循环为遍历该列的每一个元素。

这样得到的是排序好的y的list

 public class Solution {
     public int minTotalDistance(int[][] grid) {
         if (grid == null || grid.length < 1) {
             return 0;
         }
         int m = grid.length, n = grid[0].length;
         List<Integer> xList = new ArrayList<Integer>();
         List<Integer> yList = new ArrayList<Integer>();
         for (int i = 0; i < m; i++) {
             for (int j = 0; j < n; j++) {
                 if (grid[i][j] == 1) {
                     xList.add(i);
                 }
             }
         }
         for (int j = 0; j < n; j++) {
             for (int i = 0; i < m; i++) {
                 if (grid[i][j] == 1) {
                     yList.add(j);
                 }
             }
         }

         return calcTotalDistance(xList) + calcTotalDistance(yList);
     }

     private int calcTotalDistance(List<Integer> list) {
         if (list == null || list.size() < 2) {
             return 0;
         }
         int len = list.size();
         int i = 0, j = len - 1;
         int result = 0;
         while (i < j) {
             result += (list.get(j--) - list.get(i++));
         }
         return result;
     }
 }