double check 912. Best Meeting Point

Description

中文English

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|.

Have you met this question in a real interview?  

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.

这道题让我们求最佳的开会地点，该地点需要到每个为1的点的曼哈顿距离之和最小，题目中给了我们提示，让我们先从一维的情况来分析，那么我们先看一维时有两个点A和B的情况,

______A_____P_______B_______

那么我们可以发现，只要开会为位置P在[A, B]区间内，不管在哪，距离之和都是A和B之间的距离，如果P不在[A, B]之间，那么距离之和就会大于A和B之间的距离，那么我们现在再加两个点C和D：

______C_____A_____P_______B______D______

class Solution { public: /** * @param grid: a 2D grid * @return: the minimize travel distance */ //int minTotalDistance(vector<vector<int>> &grid) { // Write your code here // 这个题目是803. Shortest Distance from All Buildings的简化版本,没有障碍物。可以用greedy + math的方式解决。但不好想。 // 803用BFS int minTotalDistance(vector<vector<int>>& grid) { vector<int> rows, cols; for (int i = 0; i < grid.size(); ++i) { for (int j = 0; j < grid[i].size(); ++j) { if (grid[i][j] == 1) { rows.push_back(i); cols.push_back(j); } } } sort(cols.begin(), cols.end()); // row 已经排好序了 int res = 0, i = 0, j = rows.size() - 1; while (i < j) { res += rows[j] - rows[i] + cols[j--] - cols[i++]; } return res; } };