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:

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>& grid) {

vector 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 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>& grid) {

vector 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

 

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