[LintCode/LeetCode] Best Meeting Point

最新推荐文章于 2021-12-26 14:16:03 发布

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最新推荐文章于 2021-12-26 14:16:03 发布

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CC 4.0 BY-SA版权

原文链接：https://segmentfault.com/a/1190000015402163

本文介绍了一个给定点集合中寻找一个位置使得所有点到该位置的曼哈顿距离之和最小的问题，并提供了一种有效的解决方案。该问题适用于如物流、设施布局等实际应用场景。

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Problem

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

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.

Solution

public class Solution {
    /**
     * @param grid: a 2D grid
     * @return: the minimize travel distance
     */
    public int minTotalDistance(int[][] grid) {
        // Write your code here
        List<Integer> x = new ArrayList<>();
        List<Integer> y = new ArrayList<>();
        for (int i = 0; i < grid.length; i++) {
            for (int j = 0; j < grid[0].length; j++) {
                if (grid[i][j] == 1) {
                    x.add(i);
                    y.add(j);
                }
            }
        }
        return getMD(x) + getMD(y);
    }
    public int getMD(List<Integer> nums) {
        // zhong dian is here
        Collections.sort(nums);
        int i = 0, j = nums.size()-1;
        int distance = 0;
        while (i < j) {
            distance += nums.get(j--) - nums.get(i++);
        }
        return distance;
    }
}