296. Best Meeting Point

 

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.

Hint:

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

 

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,题目中给了提示,让从一维的情况来分析,先看一维时有两个点A和B的情况,

______A_____P_______B_______

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,现在再加两个点C和D:

______C_____A_____P_______B______D______

通过分析可以得出,P点的最佳位置就是在 [A, B] 区间内,这样和四个点的距离之和为AB距离加上 CD 距离,在其他任意一点的距离都会大于这个距离,那么分析出来了上述规律,这题就变得很容易了,只要给位置排好序,然后用最后一个坐标减去第一个坐标,即 CD 距离,倒数第二个坐标减去第二个坐标,即 AB 距离,以此类推,直到最中间停止,那么一维的情况分析出来了,二维的情况就是两个一维相加即可,参见代码如下:

 

解法一:

class Solution {
public:
    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);
                }
            }
        }
        return minTotalDistance(rows) + minTotalDistance(cols);
    }
    int minTotalDistance(vector<int> v) {
        int res = 0;
        sort(v.begin(), v.end());
        int i = 0, j = v.size() - 1;
        while (i < j) res += v[j--] - v[i++];
        return res;
    }
};

 

我们也可以不用多写一个函数,直接对 rows 和 cols 同时处理,稍稍能简化些代码:

 

解法二:

class Solution {
public:
    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());
        int res = 0, i = 0, j = rows.size() - 1;
        while (i < j) res += rows[j] - rows[i] + cols[j--] - cols[i++];
        return res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/296

 

类似题目:

Minimum Moves to Equal Array Elements II

Shortest Distance from All Buildings

 

参考资料:

https://leetcode.com/problems/best-meeting-point/

https://leetcode.com/problems/best-meeting-point/discuss/74186/14ms-java-solution

https://leetcode.com/problems/best-meeting-point/discuss/74244/Simple-Java-code-without-sorting.

https://leetcode.com/problems/best-meeting-point/discuss/74193/Java-2msPython-40ms-two-pointers-solution-no-median-no-sort-with-explanation

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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