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?
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。
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______
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。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
C++:
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; } };
C++:
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; } };
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