LeetCode 296. Best Meeting Point
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原题链接在这里:https://leetcode.com/problems/best-meeting-point/
题目:
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.
题解:
When trying to minimize the manhattan distance, it is trying to minimize the absolute deviations of x and y.
And median minimize the absolute deviations.
We get all x and y when there is a building.
And calculate absolute deviations.
Time Complexity: O(m * n). m = grid.length. n = grid[0].length.
Space: O(m + n).
AC Java:
1 class Solution { 2 public int minTotalDistance(int[][] grid) { 3 if(grid == null || grid.length == 0 || grid[0].length == 0){ 4 return 0; 5 } 6 7 int m = grid.length; 8 int n = grid[0].length; 9 10 List<Integer> iIndexList = new ArrayList<>(); 11 for(int i = 0; i < m ; i++){ 12 for(int j = 0; j < n; j++){ 13 if(grid[i][j] == 1){ 14 iIndexList.add(i); 15 } 16 } 17 } 18 19 List<Integer> jIndexList = new ArrayList<>(); 20 for(int j = 0; j < n; j++){ 21 for(int i = 0; i < m; i++){ 22 if(grid[i][j] == 1){ 23 jIndexList.add(j); 24 } 25 } 26 } 27 28 int iDist = dist(iIndexList); 29 int jDist = dist(jIndexList); 30 return iDist + jDist; 31 } 32 33 private int dist(List<Integer> list){ 34 int l = 0; 35 int r = list.size() - 1; 36 int res = 0; 37 38 while(l < r){ 39 res += list.get(r--) - list.get(l++); 40 } 41 42 return res; 43 } 44 }
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