[LeetCode 218] 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).

](https://leetcode.com/static/images/problemset/skyline1.jpg)

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

1. 如果每个rectangle都去处理就非常难处理，所以需要先预处理，将每个building拆分成左、右两个顶点。区别左右顶点的方法是：左顶点的height设为负， 右顶点height设为正。（左右顶点的处理方式不同）
   例如[ [2 9 10], [3 7 15], [5 12 12] ] ====》拆分成[[2, -10], [9, 10], [3, -15], [7, 15],[5, -12], [12, 12]]
2. 再对拆分后的cornerList按照X坐标排序（升序），如果X坐标相同，则height排升序（因为height小的是会被更高的给覆盖掉）
3. 用一个MaxHeap来维护当前扫描的rectangle的最高高度。如果该点在MaxHeap没有被移除，说明当前建筑还没有结束。
   1）左顶点：遇到就将height加入到MaxHeap
   2）右顶点：找出 MaxHeap 中对应的左顶点的height，移除。说明当前建筑结束。
   3）过程中需记录上prev MaxHeap top element。同时得到当前 MaxHeap top element
   如果 prev Top != current Top，说明
   （1）要么新加入的起点height比以前的更高，那么一定是一个拐点，加入result
   （2）或者，终点移除后，发现有比他矮的建筑，也是拐点，加入result。

class Solution {
    public List<List<Integer>> getSkyline(int[][] buildings) {
        List<List<Integer>> result = new ArrayList<> ();
        if (buildings == null || buildings.length == 0 || buildings[0].length == 0) {
            return result;
        }
        
        
        //1. Split each rectangle into left upper corner and right upper corner
        List<int[]> cornerList = new ArrayList<> ();
        for (int[] building : buildings) {
            cornerList.add (new int[] { building[0], -building[2] });
            cornerList.add (new int[] { building[1], building[2] });
        }
        
        
        //2. Sort cornerList based on x axis
        Collections.sort (cornerList, new Comparator <int[]> () {
            public int compare (int[] list1, int[] list2) {
                if (list1[0] != list2[0]) {
                    return list1[0] - list2[0];
                } else {
                    return list1[1] - list2[1];
                }
            }
        });
        
        PriorityQueue <Integer> heightMaxHeap = new PriorityQueue <> (new Comparator <Integer> () {
            public int compare (Integer i1, Integer i2) {
                return i2 - i1;
            }
        });
        
        int prevHeight = 0;
        heightMaxHeap.offer (0); // add ground horizontal line
        
        //3. scan corner list and get the result
        for (int[] corner : cornerList) {
            // left corner: push height into maxHeap, right Corner, remove the height (current building completes)
            if (corner[1] < 0) {
                heightMaxHeap.offer (-corner[1]);
            } else {
                heightMaxHeap.remove (corner[1]);
            }
            
            int currentMaxHeight = heightMaxHeap.peek ();
            if (currentMaxHeight != prevHeight) {
                prevHeight = currentMaxHeight;
                // List<Integer> temp = new ArrayList<> ();
                // temp.add (corner[0]);
                // temp.add (currentMaxHeight);
                result.add (Arrays.asList (corner[0], currentMaxHeight));
            }
            
        }
        
        return result;
    }
}