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:
- 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<Integer> xList = new ArrayList<Integer>();
8 List<Integer> yList = new ArrayList<Integer>();
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<Integer> 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<Integer> xList = new ArrayList<Integer>();
8 List<Integer> yList = new ArrayList<Integer>();
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<Integer> 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 }