leetcode: Best Meeting Point

转自：http://blog.csdn.net/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:

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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class Solution {

public:

int minTotalDistance(vector<vector<int>>& grid) {

if(grid.size() == 0)

return 0;

vector<int> rows, cols;

for(int i = 0; i < grid.size(); i++) {

for(int j = 0; j < grid[0].size(); j++) {

if(grid[i][j] == 1)

rows.push_back(i);

}

}

for(int j = 0; j < grid[0].size(); j++) {

for(int i = 0; i < grid.size(); i++) {

if(grid[i][j] == 1)

cols.push_back(j);

}

}

int distance = 0;

int mid = rows.size() / 2;

for(int i = 0; i < rows.size(); i++)

distance += abs(rows[i] - rows[mid]);

mid = cols.size() / 2;

for(int i = 0; i < cols.size(); i++)

distance += abs(cols[i] - cols[mid]);

return distance;

}

};