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 , 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))

1 public class Solution { 2     public int minTotalDistance(int[][] grid) { 3         if (grid == null || grid.length < 1) { 4             return 0; 5         } 6         int m = grid.length, n = grid[0].length; 7         List
     
       xList = new ArrayList
      
       (); 8         List
       
         yList = new ArrayList
        
         (); 9         for (int i = 0; i < m; i++) {10             for (int j = 0; j < n; j++) {11                 if (grid[i][j] == 1) {12                     xList.add(i);13                     yList.add(j);14                 }15             }16         }17         return calcTotalDistance(xList) + calcTotalDistance(yList);18     }19     20     private int calcTotalDistance(List
         
           list) {21         if (list == null || list.size() < 2) {22             return 0;23         }24         Collections.sort(list);25         int len = list.size();26         int i = 0, j = len - 1;27         int result = 0;28         while (i < j) {29             int left = list.get(i);30             int right = list.get(j);31             result += (right - left);32             i++;33             j--;34         }35         return result;36     }37 }
         
        
       
      
     

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

1 public class Solution { 2     public int minTotalDistance(int[][] grid) { 3         if (grid == null || grid.length < 1) { 4             return 0; 5         } 6         int m = grid.length, n = grid[0].length; 7         List
     
       xList = new ArrayList
      
       (); 8         List
       
         yList = new ArrayList
        
         (); 9         for (int i = 0; i < m; i++) {10             for (int j = 0; j < n; j++) {11                 if (grid[i][j] == 1) {12                     xList.add(i);13                 }14             }15         }16         for (int j = 0; j < n; j++) {17             for (int i = 0; i < m; i++) {18                 if (grid[i][j] == 1) {19                     yList.add(j);20                 }21             }22         }23         24         return calcTotalDistance(xList) + calcTotalDistance(yList);25     }26     27     private int calcTotalDistance(List
         
           list) {28         if (list == null || list.size() < 2) {29             return 0;30         }31         int len = list.size();32         int i = 0, j = len - 1;33         int result = 0;34         while (i < j) {35             result += (list.get(j--) - list.get(i++));36         }37         return result;38     }39 }