218. 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 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], ...]

Solution:

It is so hard to come out a good solution:

Here is a good reading material: https://briangordon.github.io/2014/08/the-skyline-problem.html

The idea of my solution is keep track the current maximum height of every building's left and right point. 

eg. at x=2, the maximum height=10, so the skyline is [2,10], at x=3, the maximum height=15, so add skyline[3,15], at x=5, the maximum height=15 which is equal to previous height, so we skip, at x=7 the maximum height is 12, we add [7,12], and so on.

We use a priority queue to track the maximum height. We need a way to track when is a building entering our site and exiting our site, so we can update the priority queue. In my case, I split the original building array to [x, -h] and [y,h], so when h is negative we add it to the priority queue, while if h is positive, we know that means the building is exiting our view, so we remove it from the priority queue. 

 public List<int[]> getSkyline(int[][] buildings) {  
    int n=buildings.length;  
    int[][] skyBuildings=new int[2*n][2];  
    for(int i=0;i<n;i++){  
      skyBuildings[2*i][0]=buildings[i][0];  
      skyBuildings[2*i][1]=-buildings[i][2];  
       skyBuildings[2*i+1][0]=buildings[i][1];  
      skyBuildings[2*i+1][1]=buildings[i][2];  
    }  
    Comparator<int[]> arrCompare=new Comparator<int[]>(){  
      @Override  
      public int compare(int[] a, int[] b){  
        if(a[0]==b[0]) return (a[1]-b[1]);  
        else return a[0]-b[0];  
      }  
    };  
    List<int[]> res=new ArrayList<int[]>();  
    Arrays.sort(skyBuildings,arrCompare);  
    int preH=0;  
    PriorityQueue<Integer> pq=new PriorityQueue<Integer>();  
    for(int i=0;i<2*n;i++){  
      if(skyBuildings[i][1]<0) pq.offer(skyBuildings[i][1]);  
      else pq.remove(-skyBuildings[i][1]);  
      int h=pq.size()==0? 0: pq.peek();  
      if(h!=preH){  
        int[] item=new int[]{skyBuildings[i][0],-h};  
        res.add(item);  
        preH=h;  
      }       
    }  
     return res;  
   }