The Skyline Problem

Description

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).

Buildings Skyline Contour 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], ...]

Hint

https://briangordon.github.io/2014/08/the-skyline-problem.html

Train of Thought

Now that we’re able to scan through the critical points and consider only the “active” set of rectangles at each critical point, an interesting opportunity presents itself. Our current solution can be written as: for each critical point c c.y gets the height of the tallest rectangle over c This is no longer obviously (n2)O(n2). If we can somehow calculate the height of the tallest rectangle over c in faster than (n)O(n) time, we have beaten our (n2)O(n2) algorithm. Fortunately, we know about a data structure which can keep track of an active set of integer-keyed objects and return the highest one in (logn)O(log⁡n) time: a heap. Our final solution, then, in (nlogn)O(nlog⁡n) time, is as follows. First, sort the critical points. Then scan across the critical points from left to right. When we encounter the left edge of a rectangle, we add that rectangle to the heap with its height as the key. When we encounter the right edge of a rectangle, we remove that rectangle from the heap. (This requires keeping external pointers into the heap.) Finally, any time we encounter a critical point, after updating the heap we set the height of that critical point to the value peeked from the top of the heap.

用正负来表示起点和终点

Code

public class Solution {
public List<int[]> getSkyline(int[][] buildings) {
    List<int[]> points = new ArrayList<>();
    for(int [] b : buildings){
        points.add(new int[]{b[0], -b[2]});
        points.add(new int[]{b[1], b[2]});
    }
    Collections.sort(points, new Comparator<int[]>(){
        @Override
        public int compare(int [] p1, int [] p2){
            if(p1[0] == p2[0]) return p1[1] -p2[1];
            else return p1[0] - p2[0];
        }
    });
    List<int[]> result = new ArrayList<>();
    TreeMap<Integer, Integer> treeMap = new TreeMap<>();// using TreeMap to archive TreeMultiSet
    treeMap.put(0, 1);//add a sentinel to avoid processing with empty TreeMap
    int prev = 0;
    for(int [] point: points){
        if(point[1]<0) treeMap.put(-point[1], treeMap.getOrDefault(-point[1], 0) + 1);//reach a new rectangle, height count+1
        else {
            treeMap.put(point[1], treeMap.getOrDefault(point[1], 0) - 1);//leave a new rectangle, height count-1
            if(treeMap.get(point[1]) == 0) treeMap.remove(point[1]);
        }
        int cur = treeMap.lastKey();
        if(prev != cur){//find a new skyline strip
            result.add(new int[]{point[0], cur});
            prev = cur;
        }
    }
    return result;
}

}

Complexity