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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 are given 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:
[0, 10000].Li. [...[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], ...]Points : skyline mean set of those points who change hight
1 /* 2 * Solution 1 Heap -- PriorityQueue 3 */ 4 public class Solution { 5 public List<int[]> getSkyline(int[][] buildings) { 6 List<int[]> list = new LinkedList<int[]>(); 7 List<int[]> height = new LinkedList<int[]>(); 8 9 for (int[] b : buildings) { 10 height.add(new int[]{b[0], -b[2]}); 11 height.add(new int[]{b[1], b[2]}); 12 } 13 14 Collections.sort(height, (a, b) -> { 15 if (a[0] != b[0]) 16 return a[0] - b[0]; // small to large 17 return a[1] - b[1]; // low to high 18 }); 19 20 PriorityQueue<Integer> pq = new PriorityQueue<Integer>((a, b) -> (b - a)); 21 int pre = 0; 22 pq.offer(0); 23 24 for (int[] h : height) { 25 if (h[1] < 0) { 26 pq.offer(-h[1]); 27 } else { 28 pq.remove(h[1]); 29 } 30 int cur = pq.peek(); 31 if (pre != cur) { // change of height point 32 list.add(new int[]{h[0], cur}); 33 pre = cur; 34 } 35 } 36 return list; 37 } 38 }
1 /* 2 * Solution 2 Divide and Conquer -- merge 3 */ 4 public class Solution { 5 public List<int[]> getSkyline(int[][] buildings) { 6 if (buildings.length == 0) 7 return new LinkedList<int[]>(); 8 return recurSkyline(buildings, 0, buildings.length - 1); 9 } 10 11 private LinkedList<int[]> recurSkyline(int[][] buildings, int s, int t) { 12 if (s == t) { 13 LinkedList<int[]> ret = new LinkedList<int[]>(); 14 ret.add(new int[]{buildings[s][0], buildings[s][2]}); // left top 15 ret.add(new int[]{buildings[s][1], 0}); // right, bottom 16 return ret; 17 } else { 18 int mid = (t - s) / 2 + s; 19 return merge(recurSkyline(buildings, s, mid), recurSkyline(buildings, mid + 1, t)); 20 } 21 } 22 23 private LinkedList<int[]> merge(LinkedList<int[]> l1, LinkedList<int[]> l2) { 24 int h = 0; 25 int x = 0; 26 int h1 = 0, h2 = 0; 27 LinkedList<int[]> ret = new LinkedList<int[]>(); 28 29 while (!l1.isEmpty() && !l2.isEmpty()) { 30 if (l1.getFirst()[0] < l2.getFirst()[0]) { 31 x = l1.getFirst()[0]; // getFirst, removeFirst, getLast based on LinkedList, List do not have these functions 32 h1 = l1.getFirst()[1]; 33 h = Math.max(h1, h2); 34 l1.removeFirst(); 35 } else if (l1.getFirst()[0] > l2.getFirst()[0]) { 36 x = l2.getFirst()[0]; 37 h2 = l2.getFirst()[1]; 38 h = Math.max(h1, h2); 39 l2.removeFirst(); 40 } else { 41 x = l1.getFirst()[0]; 42 h1 = l1.getFirst()[1]; 43 h2 = l2.getFirst()[1]; 44 h = Math.max(h1, h2); 45 l1.removeFirst(); 46 l2.removeFirst(); 47 } 48 49 if (ret.size() == 0 || ret.getLast()[1] != h) { // height change point 50 ret.add(new int[]{x, h}); 51 } 52 } 53 54 ret.addAll(l1); 55 ret.addAll(l2); 56 return ret; 57 } 58 }
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原文地址：http://www.cnblogs.com/joycelee/p/5468122.html
