摘要:为了尽量保证右边的点向左走,左边的点向右走,那我们就应该去这些点中间的点作为交点。由于是曼哈顿距离,我们可以分开计算横坐标和纵坐标,结果是一样的。
Best Meeting Point
横纵分离 复杂度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 - 0The point (0,2) is an ideal meeting point, as the total travel
distance of 2+2+2=6 is minimal. So return 6.
时间 O(NM) 空间 O(NM)
思路为了保证总长度最小,我们只要保证每条路径尽量不要重复就行了,比如1->2->3<-4这种一维的情况,如果起点是1,2和4,那2->3和1->2->3这两条路径就有重复了。为了尽量保证右边的点向左走,左边的点向右走,那我们就应该去这些点中间的点作为交点。由于是曼哈顿距离,我们可以分开计算横坐标和纵坐标,结果是一样的。所以我们算出各个横坐标到中点横坐标的距离,加上各个纵坐标到中点纵坐标的距离,就是结果了。
代码public class Solution { public int minTotalDistance(int[][] grid) { Listipos = new ArrayList (); List jpos = new ArrayList (); // 统计出有哪些横纵坐标 for(int i = 0; i < grid.length; i++){ for(int j = 0; j < grid[0].length; j++){ if(grid[i][j] == 1){ ipos.add(i); jpos.add(j); } } } int sum = 0; // 计算纵坐标到纵坐标中点的距离,这里不需要排序,因为之前统计时是按照i的顺序 for(Integer pos : ipos){ sum += Math.abs(pos - ipos.get(ipos.size() / 2)); } // 计算横坐标到横坐标中点的距离,这里需要排序,因为统计不是按照j的顺序 Collections.sort(jpos); for(Integer pos : jpos){ sum += Math.abs(pos - jpos.get(jpos.size() / 2)); } return sum; } }
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Problem 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 ...
Problem 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 ...
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