218. The Skyline Problem
题目：
A city's skyline is the outer contour of the silhouette formed by all the buildings in that city when viewed from a distance. Now suppose you aregiven the locations and height of all the buildings as shown on a cityscape photo (Figure A), write a program to output the skyline formed by these buildings collectively (Figure B).
The geometric information of each building is represented by a triplet of integers [Li, Ri, Hi], where Li and Ri are the x coordinates of the left and right edge of the ith building, respectively, and Hi is its height. It is guaranteed that 0 ≤ Li, Ri ≤ INT_MAX, 0 < Hi ≤ INT_MAX, and Ri - Li > 0. You may assume all buildings are perfect rectangles grounded on an absolutely flat surface at height 0.
For instance, the dimensions of all buildings in Figure A are recorded as: [ [2 9 10], [3 7 15], [5 12 12], [15 20 10], [19 24 8] ] .
The output is a list of "key points" (red dots in Figure B) in the format of [ [x1,y1], [x2, y2], [x3, y3], ... ] that uniquely defines a skyline. A key point is the left endpoint of a horizontal line segment. Note that the last key point, where the rightmost building ends, is merely used to mark the termination of the skyline, and always has zero height. Also, the ground in between any two adjacent buildings should be considered part of the skyline contour.
For instance, the skyline in Figure B should be represented as:[ [2 10], [3 15], [7 12], [12 0], [15 10], [20 8], [24, 0] ].
Notes:
- The number of buildings in any input list is guaranteed to be in the range
[0, 10000]. - The input list is already sorted in ascending order by the left x position
Li. - The output list must be sorted by the x position.
- There must be no consecutive horizontal lines of equal height in the output skyline. For instance,
[...[2 3], [4 5], [7 5], [11 5], [12 7]...]is not acceptable; the three lines of height 5 should be merged into one in the final output as such:[...[2 3], [4 5], [12 7], ...]
链接： http://leetcode.com/problems/the-skyline-problem/
题解：
很经典的题目。一开始的想法是遍历一遍数组，然后把所有关键点及其长度计算出来，再遍历一遍数组，找到所有结果。没有尝试。
后来研究了Discuss，发现有两种做法：
- 一种是用Mergesort的原理。
- 首先base case是当我们递归函数的lo == hi时，这时只有一个building，由一个building形成一个skyline。这里我们的两个关键点是(left, height)以及(right，0)。
- 有了skyline之后我们就可以进行推广。每个skyline都已经是一个left sorted的doubly linkedlist。 每次由两个skyline merge成一个更大的skyline，最后一步步得出结果。 这里merge函数跟"Merge Two Sorted List" 很象。要注意merge时的各种判断，我们需要current height1和current height2来cache skyline1和skyline2的当前高度。而最后加入结果的height为Math.max(skyline1 current height，skyline2 current height)，加入结果的index为两个skyline元素中最左边的那一个，之后还要把处理过的元素从skyline1或者2，或者1和2里去除掉。所以这里用doubly-linked list很方便，其实用queue或者deque也能有一样的效果。
- merge是一个O(n)的操作，而总算法的时间复杂度是O(nlogn)，空间复杂度是O(n)。
- 另外一种是维护一个heap，用Sweepling Algorithm在heap里面增删改，最后输出结果。
Divide and Conquer:
Time Complexity - O(nlogn)， Space Complexity - O(n)
public class Solution { public List<int[]> getSkyline(int[][] buildings) { if(buildings == null || buildings.length == 0) return new LinkedList<int[]>(); return getSkyline(buildings, 0, buildings.length - 1); } private LinkedList<int[]> getSkyline(int[][] buildings, int lo, int hi) { if(lo < hi) { int mid = lo + (hi - lo) / 2; return mergeSkylines(getSkyline(buildings, lo, mid), getSkyline(buildings, mid + 1, hi)); } else { // lo == hi, base case, add one building to skyline LinkedList<int[]> skyline = new LinkedList<int[]>(); skyline.add(new int[]{buildings[lo][0], buildings[lo][2]}); skyline.add(new int[]{buildings[lo][1], 0}); return skyline; } } private LinkedList<int[]> mergeSkylines(LinkedList<int[]> skyline1, LinkedList<int[]> skyline2) { // merge two Skylines LinkedList<int[]> skyline = new LinkedList<int[]>(); int height1 = 0, height2 = 0; while(skyline1.size() > 0 && skyline2.size() > 0) { int index = 0, height = 0; if(skyline1.getFirst()[0] < skyline2.getFirst()[0]) { index = skyline1.getFirst()[0]; height1 = skyline1.getFirst()[1]; height = Math.max(height1, height2); skyline1.removeFirst(); } else if (skyline1.getFirst()[0] > skyline2.getFirst()[0]) { index = skyline2.getFirst()[0]; height2 = skyline2.getFirst()[1]; height = Math.max(height1, height2); skyline2.removeFirst(); } else { index = skyline1.getFirst()[0]; height1 = skyline1.getFirst()[1]; height2 = skyline2.getFirst()[1]; height = Math.max(height1, height2); skyline1.removeFirst(); skyline2.removeFirst(); } if(skyline.size() == 0 || height != skyline.getLast()[1]) skyline.add(new int[]{index, height}); } skyline.addAll(skyline1); skyline.addAll(skyline2); return skyline; } }
Sweeping line + Heap: (二刷再解决)
二刷:
二刷依然是使用了Divide and Conquer。使用了ArrayList来保存结果，虽然代码长一点，但速度更快，并且更容易理解一点点。
要注意的还是在merge的时候：
- 我们仍然要保存之前的height1和height2两个变量。
- 对skyline1和skyline2中的第一个点p1和p2，我们分三种情况
- p1[0] < p2[0]，这时候p1在p2之前出现，我们更新height1 = p1[1]，先处理p1
- p1[0] > p2[0]，这时候p1在p2之后出现，我们更新height2 = p2[1]，先处理p2
- 否则两点同时出现，我们更新height1和height2，同时处理两个点
- 在height = max(height1, height2)，并且height != res.get(res.size() - 1)时，也就是当前这个点的高度不等于之前点的高度，我们把这个点加入到结果集res中。这里的一个小边界条件是当res.size() == 0时，我们也加入这个点[index, height]。
- 在merge的最后我们要判断一下是否skyline1或者skyline2中所有的点都计算过了。 假如还有剩余点，我们对其进行处理。
也可以用一些特殊的数据结构来做，比如PQ + Sweeping line， Treap， Fenwick Tree等等。 下次再研究。
Java:
Time Complexity - O(nlogn)， Space Complexity - O(n)
public class Solution { public List<int[]> getSkyline(int[][] buildings) { if (buildings == null || buildings.length == 0) return new ArrayList<int[]>(); return getSkyline(buildings, 0, buildings.length - 1); } private List<int[]> getSkyline(int[][] buildings, int lo, int hi) { if (lo < hi) { int mid = lo + (hi - lo) / 2; return mergeSkylines(getSkyline(buildings, lo, mid), getSkyline(buildings, mid + 1, hi)); } else { List<int[]> res = new ArrayList<>(); res.add(new int[] {buildings[lo][0], buildings[lo][2]}); res.add(new int[] {buildings[lo][1], 0}); return res; } } private List<int[]> mergeSkylines(List<int[]> skyline1, List<int[]> skyline2) { List<int[]> res = new ArrayList<>(); int i = 0, j = 0; int index = 0, height = 0, height1 = 0, height2 = 0; while (i < skyline1.size() && j < skyline2.size()) { int[] p1 = skyline1.get(i); int[] p2 = skyline2.get(j); if (p1[0] < p2[0]) { index = p1[0]; height1 = p1[1]; i++; } else if (p1[0] > p2[0]) { index = p2[0]; height2 = p2[1]; j++; } else { index = p1[0]; height1 = p1[1]; height2 = p2[1]; i++; j++; } height = Math.max(height1, height2); if (res.size() == 0 || height != res.get(res.size() - 1)[1]) res.add(new int[] {index, height}); } if (i < skyline1.size()) { for (int k = i; k < skyline1.size(); k++) { int[] p1 = skyline1.get(k); if (p1[1] != res.get(res.size() - 1)[1]) res.add(p1); } } else if (j < skyline2.size()){ for (int k = j; k < skyline2.size(); k++) { int[] p2 = skyline2.get(k); if (p2[1] != res.get(res.size() - 1)[1]) res.add(p2); } } return res; } }
Reference:
https://en.wikipedia.org/wiki/Sweep_line_algorithm#Applications
http://www.algorithmist.com/index.php/UVa_105
http://www.cnblogs.com/easonliu/p/4531020.html
https://cseweb.ucsd.edu/classes/sp04/cse101/skyline.pdf
http://sandrasi-sw.blogspot.com/2012/12/the-skyline-problem.html
http://www.geeksforgeeks.org/divide-and-conquer-set-7-the-skyline-problem/
https://briangordon.github.io/2014/08/the-skyline-problem.html
https://leetcode.com/discuss/61274/17-line-log-time-space-accepted-easy-solution-explanations
https://leetcode.com/discuss/37630/my-c-code-using-one-priority-queue-812-ms
https://leetcode.com/discuss/37736/108-ms-17-lines-body-explained
https://leetcode.com/discuss/40963/share-my-divide-and-conquer-java-solution-464-ms
https://leetcode.com/discuss/54201/short-java-solution
https://leetcode.com/discuss/88149/java-solution-using-priority-queue-and-sweepline
