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|
.
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:
链接: http://leetcode.com/problems/best-meeting-point/
题解:
很有意思的一道题目,假设二维数组中一个点到其他给定点的Manhattan Distance最小,求distance和。 因为在一维数组中这个distance最小的点就是给定所有点的median,题目又给定使用曼哈顿距离,我们就可以把二维计算分解成为两个一维的计算。应该还可以用DP的方法解决,判断用哪一种方法其实非常复杂,依赖于mn和排序的比较。
Time Complexity - Math.max(O(mn), O(klogk)), Space Complexity - O(m + n)。
public class Solution { public int minTotalDistance(int[][] grid) { List<Integer> xAxis = new ArrayList<>(); List<Integer> yAxis = new ArrayList<>(); for(int i = 0; i < grid.length; i++) { for(int j = 0; j < grid[0].length; j++) { if(grid[i][j] == 1) { xAxis.add(i); yAxis.add(j); } } } return getMin(xAxis) + getMin(yAxis); } private int getMin(List<Integer> list) { Collections.sort(list); int res = 0; int lo = 0, hi = list.size() - 1; while(lo < hi) { res += list.get(hi--) - list.get(lo++); // hi - mid + mid - lo = hi - lo } return res; } }
Reference:
https://leetcode.com/discuss/65336/14ms-java-solution
https://leetcode.com/discuss/65366/o-mn-java-2ms
https://leetcode.com/discuss/65464/java-python-40ms-pointers-solution-median-sort-explanation
https://leetcode.com/discuss/66401/the-only-person-dont-know-median-could-give-shortest-distance
http://math.stackexchange.com/questions/113270/the-median-minimizes-the-sum-of-absolute-deviations
https://leetcode.com/discuss/65510/simple-java-code-without-sorting
http://www.jiuzhang.com/problem/30/
原文:http://www.cnblogs.com/yrbbest/p/5047006.html