296 Best Meeting Point

1. Question

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):
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1 - 0 - 0 - 0 - 1
2
| | | | |
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0 - 0 - 0 - 0 - 0
4
| | | | |
5
0 - 0 - 1 - 0 - 0
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The point(0,2)is an ideal meeting point, as the total travel distance of 2+2+2=6 is minimal. So return 6

2. Implementation

(1) Sort + Two Pointers
思路: 和#462的思路一样, 相遇的点在哪不影响最后的距离
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class Solution {
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public int minTotalDistance(int[][] grid) {
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List<Integer> rows = new ArrayList<>();
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List<Integer> cols = new ArrayList<>();
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int m = grid.length, n = grid[0].length;
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for (int i = 0; i < m; i++) {
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for (int j = 0; j < n; j++) {
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if (grid[i][j] == 1) {
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rows.add(i);
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cols.add(j);
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}
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}
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}
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return getDistance(rows, false) + getDistance(cols, true);
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}
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public int getDistance(List<Integer> list, boolean isCol) {
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// rows这个list本身已经sorted,所以只对cols排序
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if (isCol) {
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Collections.sort(list);
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}
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int start = 0, end = list.size() - 1;
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int dist = 0;
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while (start < end) {
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dist += list.get(end) - list.get(start);
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++start;
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--end;
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}
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return dist;
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}
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}
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3. Time & Space Complexity

Sort + Two Pointers: 时间复杂度O(mn + mlogm + nlogn), 空间复杂度O(Max(m,n))