LeetCode 218. The Skyline Problem(java)

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], …]

思路：类似于meeting room的方法，用hashmap存《start, +1》和《end, -1》。在这里就是先用一个List《int[]》来存储《start, height》和《end, -height》, 然后把这个list按照从小到大排序，最后用一个queue来走一遍每个点，每次遇到正值的height，就offer，遇到负值就pop此值，然后在每一个点，都peek heap中的最大height，如果这个height和之前存在result里的值不同，将此点和值存进result，否则更新pre，继续向后走，直到走完。

代码：

public List<int[]> getSkyline(int[][] buildings) {
List<int[]> list = new ArrayList<>();
if (buildings.length == 0 || buildings[0].length == 0) return list;
List<int[]> height = new ArrayList<>();
for (int i = 0; i < buildings.length; i++) {
height.add(new int[]{buildings[i][0], buildings[i][2]});
height.add(new int[]{buildings[i][1], -buildings[i][2]});
}
Collections.sort(height, new Comparator<int[]>(){
public int compare(int[] o1, int[] o2) {
//注意这里是O2[1] - o1[0]!
if (o1[0] == o2[0]) return o2[1] - o1[1];
else return o1[0] - o2[0];
}
});
PriorityQueue<Integer> queue = new PriorityQueue<>(new Comparator<Integer>() {
public int compare(Integer o1, Integer o2) {
return o2 - o1;
}
});
queue.offer(0);
int cur = 0, pre = 0;
for (int[] h : height) {
if (h[1] > 0) {
queue.offer(h[1]);
} else {
queue.remove(-h[1]);
}
cur = queue.peek();
if (pre != cur) {
list.add(new int[]{h[0], cur});
pre = cur;
}
}
return list;
}