The Skyline Problem

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 skylineformed 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], ...]

The basic idea is that whenever there is a height difference, we need to record it as an outline. To implement it, we can use a list to store all defining points of each building. The list is sorted from left to right to get the order of occurrence of each building. To distinguish the left and right side, we select one as positive and one as negative. We then use a priority queue to track the heights of the buildings. When left point is seen, we add the height to the queue, when right point is seen, we know we no longer need to consider the building, so we remove the height from the queue. The top of the queue is the highest building we need to consider, so if it's different from the previous one, we are seeing a height difference, so we need to add the point to the result. Remember to add height 0 to the queue at first. This allows us to track if there is a gap between two buildings.

public List getSkyline(int[][] buildings) {
        List keyPoints = new ArrayList<>();
        List rst = new ArrayList<>();
        
        for (int[] point : buildings) {
            keyPoints.add(new int[]{point[0], -point[2]});
            keyPoints.add(new int[]{point[1], point[2]});
        }
        
        Collections.sort(keyPoints, new Comparator() {
            
           @Override
           public int compare(int[] first, int[] second) {
               if (first[0] != second[0]) {
                   return first[0] - second[0];
               } else {
                   return first[1] - second[1];
               }
           } 
        });
        
        int prev = -1;

        PriorityQueue heights = new PriorityQueue<>(new Comparator() {
            
            @Override
            public int compare(Integer a, Integer b) {
                return b - a;
            }
            
        });
        heights.add(0);
        for(int[] point : keyPoints) {
            if (point[1] < 0) {
                heights.add(-point[1]);
            } else {
                heights.remove(point[1]);
            }
            int curr = heights.peek();
            if (curr != prev) {
               rst.add(new int[]{point[0], curr});
                prev = curr;
            }
        }
        return rst;
    }