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], ...]
Thinking:
(1) 自建一个名为Height的数据结构，保存一个building的index和height。约定，当height为负数时表示这个高度为height的building起始于index；height为正时表示这个高度为height的building终止于index。
(2) 对building数组进行处理，每一行[ Li, Ri, Hi ]，根据Height的定义，转换为两个Height的对象，即，Height(Li, -Hi) 和 Height(Ri, Hi)。 将这两个对象存入heights这个List中。
(3) 写个Comparator对heights进行升序排序，首先按照index的大小排序，若index相等，则按height大小排序，以保证一栋建筑物的起始节点一定在终止节点之前。
(4) 将heights转换为结果。使用PriorityQueue对高度值进行暂存。遍历heights，遇到高度为负值的对象时，表示建筑物的起始节点，此时应将这个高度加入PriorityQueue。遇到高度为正值的对象时，表示建筑物的终止节点，此时应将这个高度从PriorityQueue中除去。且在遍历的过程中检查，当前的PriorityQueue的peek()是否与上一个iteration的peek()值（prev）相同，若否，则应在结果中加入[当前对象的index, 当前PriorityQueue的peek()]，并更新prev的值。
思路是照抄这个链接的：http://www.cnblogs.com/easonliu/p/4531020.html
（脑洞：C++全忘光了，这个链接的代码好久才看懂，赶快补！）
Java Code:
public class Solution {
public List<int[]> getSkyline(int[][] buildings) {
List<int[]> result = new ArrayList<int[]>();
if (buildings == null || buildings.length == 0 || buildings[0].length == 0) {
return result;
}
List<Height> heights = new ArrayList<Height>();
for (int[] building : buildings) {
heights.add(new Height(building[0], -building[2]));
heights.add(new Height(building[1], building[2]));
}
Collections.sort(heights, new Comparator<Height>() {
@Override
public int compare(Height h1, Height h2) {
return h1.index != h2.index ? h1.index - h2.index : h1.height - h2.height;
}
});
PriorityQueue<Integer> pq = new PriorityQueue<Integer>(1000, Collections.reverseOrder());
pq.offer(0);
int prev = 0;
for (Height h : heights) {
if (h.height < 0) {
pq.offer(-h.height);
} else {
pq.remove(h.height);
}
int cur = pq.peek();
if (cur != prev) {
result.add(new int[]{h.index, cur});
prev = cur;
}
}
return result;
}
class Height {
int index;
int height;
Height(int index, int height) {
this.index = index;
this.height = height;
}
}
}
