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], ...]
public class Solution {
public static class Line implements Comparable<Line> {
public int x;
public boolean left;
public int height;
public Line(int x, int height, boolean left) {
this.x = x;
this.height = height;
this.left = left;
}
@Override
public int compareTo(Line o) {
// TODO Auto-generated method stub
if (x != o.x) {
return x - o.x;
}
if (left != o.left) {
if (left) {
return -1;
} else {
return 1;
}
} else {
if (left) {
return o.height - height;
} else {
return height - o.height;
}
}
}
}
public List<int[]> getSkyline(int[][] buildings) {
List<int[]> res = new ArrayList<>();
List<Line> lines = new ArrayList<>();
for (int i = 0; i < buildings.length; i++) {
lines.add(new Line(buildings[i][0], buildings[i][2], true));
lines.add(new Line(buildings[i][1], buildings[i][2], false));
}
Collections.sort(lines);
PriorityQueue<Integer> heap = new PriorityQueue<>(11, Collections.reverseOrder());
for (Line line : lines) {
if (line.left) {
heap.add(line.height);
} else {
heap.remove(line.height);
}
int height = heap.size() == 0 ? 0 : heap.peek();
if (line.height >= height) {
int[] tmp = new int[]{line.x, height};
if (res.size() == 0) {
res.add(tmp);
} else {
int[] last = res.get(res.size()-1);
if (tmp[1] != last[1]) {
res.add(tmp);
}
}
}
}
return res;
}
}
本文探讨了如何通过编程实现城市天际线的可视化,详细解释了每个建筑的几何信息如何转换为独特的天际线轮廓。通过提供一个算法实例,包括建筑物的位置、高度和宽度,本文展示了如何输出由所有建筑物共同形成的天际线轮廓。

1247

被折叠的 条评论
为什么被折叠?



