218 The Skyline Problem

218. The Skyline Problem

1. Question

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], whereLiandRiare the x coordinates of the left and right edge of the ith building, respectively, andHiis its height. It is guaranteed that0 ≤ Li, Ri ≤ INT_MAX,0 < Hi ≤ INT_MAX, andRi - 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 positionLi.

• 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], ...]

2. Implementation

思路:

1. 将矩阵离散化，把每个building的数据分解成【左Index, 高度】和【右Index, 高度】 的形式，为了方便区分哪个index是代表矩阵的开始和末尾，我们把左Index对应的高度设为负数

2. 将分解后的数据按照Index和高度从低到高排序

3. 通过一个MaxHeap存储building的高度，遍历分解后的数据过程，如果index是负数，将其对应的高度放入MaxHeap, 否则从MaxHeap删除其对应高度

4. 维护两个变量preHeight和curHeight分别之前已知最高高度和当前最高高度，当preHeight不等于curHeight时，将对应的点加入结果中

(1) Heap(Priority Queue) + Scan Line

public class Solution {
    public List<int[]> getSkyline(int[][] buildings) {
        List<int[]> res = new ArrayList<>();

        if (buildings == null || buildings.length == 0) {
            return res;
        }

        List<int[]> points = new ArrayList<>();

        // Decompose the coordinate of a building into two ponits
        for (int[] building : buildings) {
            points.add(new int[] {building[0], -building[2]});
            points.add(new int[] {building[1], building[2]});
        }

        Collections.sort(points, (a,b) -> {
            return a[0] == b[0] ? a[1] - b[1] : a[0] - b[0];
        });


        PriorityQueue<Integer> heightMaxHeap = new PriorityQueue<>(Collections.reverseOrder());
        heightMaxHeap.add(0);

        int prevHeight = 0;
        int curHeight = 0;

        for (int[] point : points) {
            if (point[1] < 0) {
                heightMaxHeap.add(-point[1]);
            }
            else {
                heightMaxHeap.remove(point[1]);
            }

            curHeight = heightMaxHeap.peek();
            if (prevHeight != curHeight) {
                res.add(new int[] {point[0], curHeight});
                prevHeight = curHeight;
            }
        }
        return res;
    }
}

(2) Heap(TreeMap) + Scan Line

由于Priority Queue的remove()时间复杂度是O(n), 所以用TreeMap来实现heap, remove的时间复杂度是O(logn)

 class Solution {
    public List<int[]> getSkyline(int[][] buildings) {
        List<int[]> res = new ArrayList<>();

        if (buildings == null | buildings.length == 0) {
            return res;
        }

        List<int[]> points = new ArrayList<>();

        for (int[] building : buildings) {
            points.add(new int[] {building[0], -building[2]});
            points.add(new int[] {building[1], building[2]});
        }

        Collections.sort(points, new Comparator<int[]>() {
            @Override
            public int compare(int[] a, int[] b) {
                return a[0] == b[0] ? a[1] - b[1] : a[0] - b[0];
            }
        });

        TreeMap<Integer, Integer> heightMaxHeap = new TreeMap(new Comparator<Integer>() {
            @Override
            public int compare(Integer i1, Integer i2) {
                return i2 - i1;
            }
        });
        // Place Holder, 用来保证每个skyline没有overlap时，最后一点能输出. 注意观察，skyline最后一点都是的高度都是0
        heightMaxHeap.put(0, 1);

        int preHeight = 0;
        int curHeight = 0;

        for (int[] point : points) {
            if (point[1] < 0) {
                heightMaxHeap.put(-point[1], heightMaxHeap.getOrDefault(-point[1], 0) + 1);
            }
            else {
                int count = heightMaxHeap.get(point[1]);
                if (count == 1) {
                    heightMaxHeap.remove(point[1]);
                }
                else {
                    heightMaxHeap.put(point[1], count - 1);
                }
            }

            curHeight = heightMaxHeap.firstKey();

            if (preHeight != curHeight) {
                res.add(new int[] {point[0], curHeight});
                preHeight = curHeight;
            }
        }
        return res;
    }
}

3. Time & Space Complexity

Heap(TreeMap) + Scan Line: 时间复杂度O(nlogn), 空间复杂度O(n)