296. Best Meeting Point

最新推荐文章于 2021-11-27 00:46:05 发布

小菜鸟的leetcode

最新推荐文章于 2021-11-27 00:46:05 发布

阅读量921

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode

本文链接：https://blog.youkuaiyun.com/Scarlett_leetcode/article/details/50547769

leetcode 专栏收录该内容

11 篇文章

订阅专栏

本文探讨了一个二维网格中寻找一个点，使得该点到所有标记点的曼哈顿距离之和最小的问题。通过将问题简化为一维并找到中间点来解决，提供了一个高效的算法实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 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

Solution

如果是一维的，那就是以最中间的那个点为见面地点（ buggy : 不是坐标的平均值）

class Solution {
public:
    int minTotalDistance(vector<vector<int>>& grid) {
        vector<int> iArr, jArr;
        int iSum = 0, jSum = 0;
        for( int i = 0; i < grid.size(); i++){
            for( int j = 0; j < grid[0].size(); j++){
                if(grid[i][j] == 1){
                    iArr.push_back(i);
                    jArr.push_back(j);
                    iSum += i;
                    jSum += j;
                }
            }
        }
        
        const int N = iArr.size();
        sort( jArr.begin(), jArr.end());
        int iMid = iArr[N/2], jMid = jArr[N/2], rst = 0;
        
        for( int i = 0; i < N; i++){
            rst += abs( iArr[i] - iMid) + abs( jArr[i] - jMid);
        }
        return rst;
    }
};