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, 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.
Note
和到各个建筑最短距离不同，不需要暴力去遍历。这里任何点都可以是目标点，也不会有到达不了的情况
Complexity analysis
Time complexity: O(mn)
Space complexity: O(mn)
Code
public int minTotalDistance(int[][] grid) {
List<Integer> rows = collectRows(grid);
List<Integer> cols = collectCols(grid);
int row = rows.get(rows.size() / 2);
int col = cols.get(cols.size() / 2);
return minDistance1D(rows, row) + minDistance1D(cols, col);
}
private int minDistance1D(List<Integer> points, int origin) {
int distance = 0;
for (int point : points) {
distance += Math.abs(point - origin);
}
return distance;
}
private List<Integer> collectRows(int[][] grid) {
List<Integer> rows = new ArrayList<>();
for (int row = 0; row < grid.length; row++) {
for (int col = 0; col < grid[0].length; col++) {
if (grid[row][col] == 1) {
rows.add(row);
}
}
}
return rows;
}
private List<Integer> collectCols(int[][] grid) {
List<Integer> cols = new ArrayList<>();
for (int col = 0; col < grid[0].length; col++) {
for (int row = 0; row < grid.length; row++) {
if (grid[row][col] == 1) {
cols.add(col);
}
}
}
return cols;
}
更elegant的写法
public int minTotalDistance(int[][] grid) {
List<Integer> rows = collectRows(grid);
List<Integer> cols = collectCols(grid);
return minDistance1D(rows) + minDistance1D(cols);
}
private int minDistance1D(List<Integer> points) {
int distance = 0;
int i = 0;
int j = points.size() - 1;
while (i < j) {
distance += points.get(j) - points.get(i);
i++;
j--;
}
return distance;
}
