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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.
Solution 1. Median in Math. Collect coordinates in sorted order.
1 public class Solution { 2 public int minTotalDistance(int[][] grid) { 3 int m = grid.length, n = grid[0].length; 4 int total = 0, Z[] = new int[m*n]; 5 for (int dim=0; dim<2; ++dim) { 6 int i = 0, j = 0; 7 if (dim == 0) { 8 for (int x=0; x<n; ++x) 9 for (int y=0; y<m; ++y) 10 if (grid[y][x] == 1) 11 Z[j++] = x; 12 } else { 13 for (int y=0; y<m; ++y) 14 for (int g : grid[y]) 15 if (g == 1) 16 Z[j++] = y; 17 } 18 while (i < --j) 19 total += Z[j] - Z[i++]; 20 } 21 return total; 22 } 23 }
Solution 2. Count how many people live in each row and each column. Only O(m+n) space.
1 public class Solution { 2 public int minTotalDistance(int[][] grid) { 3 int m = grid.length, n = grid[0].length; 4 int[] I = new int[m], J = new int[n]; 5 for (int i=0; i<m; ++i) 6 for (int j=0; j<n; ++j) 7 if (grid[i][j] == 1) { 8 ++I[i]; 9 ++J[j]; 10 } 11 12 int total = 0; 13 for (int[] K : new int[][]{ I, J }) { 14 int i = 0, j = K.length - 1; 15 while (i < j) { 16 int k = Math.min(K[i], K[j]); 17 total += k * (j - i); 18 if ((K[i] -= k) == 0) ++i; 19 if ((K[j] -= k) == 0) --j; 20 } 21 } 22 return total; 23 } 24 }
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原文地址：http://www.cnblogs.com/joycelee/p/5274426.html
