leetcode: Best Meeting Point

转自：http://blog.youkuaiyun.com/pointbreak1/article/details/49422265

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:

1. Try to solve it in one dimension first. How can this solution apply to the two dimension case?

题解：

先考虑所有点的row值，如上图为0， 0， 2，则中值为0， 所有点距中值的距离和为(0 - 0) + (0 - 0) + (2 - 0) = 2

同理，考虑所有点的cols值，上图为，0, 2, 4，中值为2，所有点距中值的距离和为(2- 0) + (2 - 2) + (4 - 2) = 4，

所以总距离为2 + 4 = 6。

C++版：

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1. class Solution {  
2. public:  
3.     int minTotalDistance(vector<vector<int>>& grid) {  
4.         if(grid.size() == 0)  
5.             return 0;  
6.         vector<int> rows, cols;  
7.         for(int i = 0; i < grid.size(); i++) {  
8.             for(int j = 0; j < grid[0].size(); j++) {  
9.                 if(grid[i][j] == 1)  
10.                     rows.push_back(i);  
11.             }  
12.         }  
13.         for(int j = 0; j < grid[0].size(); j++) {  
14.             for(int i = 0; i < grid.size(); i++) {  
15.                 if(grid[i][j] == 1)  
16.                     cols.push_back(j);  
17.             }  
18.         }  
19.         int distance = 0;  
20.         int mid = rows.size() / 2;  
21.         for(int i = 0; i < rows.size(); i++)  
22.             distance += abs(rows[i] - rows[mid]);  
23.         mid = cols.size() / 2;  
24.         for(int i = 0; i < cols.size(); i++)  
25.             distance += abs(cols[i] - cols[mid]);  
26.         return distance;  
27.     }  
28. };