296. Best Meeting Point

问题描述：

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|.

Example:

Input: 

1 - 0 - 0 - 0 - 1
|   |   |   |   |
0 - 0 - 0 - 0 - 0
|   |   |   |   |
0 - 0 - 1 - 0 - 0

Output: 6 

Explanation: Given three people living at (0,0), (0,4), and (2,2):
             The point (0,2) is an ideal meeting point, as the total travel distance 
             of 2+2+2=6 is minimal. So return 6.

解题思路：

我的想法是，首先找到所有为1的点，用一个vector存储起来

再遍历整个矩阵并计算每个点到其他房子的距离，取最小值返回。

时间复杂度为O(mnk)其中k为为1的点。

再来看看最快的解法：

参考解释：6ms C++ one pass O(mn) solution no sort, Easiest Solution O(n) + O(m) extra space

因为我们使用的是汉密尔顿距离，而汉密尔顿距离的水平距离和垂直距离是相互独立的，并不会互相影响。

代码：

O(mnk):

class Solution {
public:
    int minTotalDistance(vector<vector<int>>& grid) {
        vector<pair<int,int>> peoples;
        int m = grid.size(), n = grid[0].size();
        for(int i = 0 ; i < m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1) peoples.push_back({i,j});
            }
        }
        int ret = INT_MAX;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                int cur = 0;
                pair<int,int> pos = {i, j};
                for(auto p : peoples){
                    cur += calDistance(p,pos);
                }
                ret = min(cur, ret);
            }
        }
        return ret;
    }
    int calDistance(pair<int,int> a, pair<int,int> b){
        return abs(a.first - b.first) + abs(a.second - b.second);
    }
};

最快的解法：

class Solution {
public:
    int findDist(const vector<int>& points) {
        int dist = 0;
        int i = 0;
        int j = points.size() - 1;
        while (i < j) {
            dist += (points[j] - points[i]);
            ++i;
            --j;
        }
        return dist;
    }
    int minTotalDistance(vector<vector<int>>& grid) {
        vector<int> rows;
        vector<int> 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 i = 0; i < grid[0].size(); ++i) {
            for (int j = 0; j < grid.size(); ++j) {
                if (grid[j][i] == 1) {
                    cols.push_back(i);
                }
            }
        }
        int ans = findDist(rows) + findDist(cols);
        return ans;
    }
};