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:

• 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], ...]

这道题真的是太难了。参考了大神博客（）才依葫芦画瓢写出来。

• 扫描线大法。将每一线段的左右节点分开，按从小到大排序。
• heap记录当前的动态高度
• 如果遇到左节点，就push到heap
• 遇到右节点就从heap中remove高度
• 对当前时间轴的操作完成后，检查和previous hight是否相同，不同的话就记录高度变化

class Solution {        class Node {        int x;        int h;        int lr;   // indicate left or right (left: -1, right: 1).        public Node(int x, int h, int lr) {            this.x = x;            this.h = h;            this.lr = lr;        }    }         class MyComparator implements Comparator
       
         {        @Override        public int compare(Node a, Node b) {            if (a.x != b.x) {                return a.x - b.x;            }                        if (a.lr == -1 && b.lr == -1) {                return b.h - a.h;            } else if (a.lr == 1 && b.lr == 1) {                return a.h - b.h;            } else {                return a.lr - b.lr;            }        }    }        public List
        
         
          > getSkyline(int[][] buildings) {        List
          
           
            > res = new ArrayList<>();                if (buildings == null || buildings.length == 0) {            return res;        }                int pre = 0;  // 以前的高度        PriorityQueue
            
              heap = new PriorityQueue<>((x, y) -> y-x); heap.offer(0); List
             
               nodes = new ArrayList<>(); for (int i = 0; i < buildings.length; i++) { nodes.add(new Node(buildings[i][0], buildings[i][2], -1)); nodes.add(new Node(buildings[i][1], buildings[i][2], 1)); } nodes.sort(new MyComparator()); for (Node n : nodes) { System.out.println(n.x + ", " + n.h + ", " + n.lr); if (n.lr == -1) { heap.offer(n.h); } else { heap.remove(n.h); } if (pre != heap.peek()) { List
              
                e = new ArrayList<>(); e.add(n.x); e.add(heap.peek()); pre = heap.peek(); res.add(e); } } return res; }}