LeetCode 296: Best Meeting Point

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最新推荐文章于 2025-05-31 02:45:00 发布

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分类专栏： LeetCode 题解 Multiple Passes

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本文介绍了一个有趣的问题：如何找到一个点使得二维网格中多个指定点到该点的曼哈顿距离之和最小，并提供了一种有效的算法解决方案。通过分别计算x轴和y轴上的最优位置，可以快速求解。

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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 usingManhattan 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.

For one dimension axises, the middle point is the shortest distance. Thus, we just need to calculate x axis and y axis seperately.

http://blog.youkuaiyun.com/xudli/article/details/49420623 Code is from here.

    public class Solution {  
        //(0,0), (0,4), and (2,2)  
        public int minTotalDistance(int[][] grid) {  
            List<Integer> x = new ArrayList<>();  
            List<Integer> y = new ArrayList<>();  
              
            for(int i=0; i<grid.length; i++) {  
                for(int j=0; j<grid[0].length; j++) {  
                    if(grid[i][j]==1) {  
                        x.add(i);  
                        y.add(j);  
                    }  
                }  
            }  
              
            return getMP(x) + getMP(y);  
        }  
          
        private int getMP(List<Integer> l) {  
            Collections.sort(l);  
            int i=0, j=l.size()-1;  
            int res = 0;  
            /*
              or set up a mid point. int mid = (0 + x.size()) / 2;
            */
            while(i<j) {  
                res += l.get(j--) - l.get(i++);  
            }  
            return res;  
        }  
    }