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LeetCode 218. The Skyline Problem 天际线问题(C++/Java)

silentteller 人气:5

题目:

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_MAX0 < 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], ...]

分析:

给定一组描述建筑物的三元组[Li, Ri, Hi],其中 Li 和 Ri 分别是第 i 座建筑物左右边缘的 x 坐标,Hi 是其高度。最后求出一组坐标来表示天际线。

利用扫描线法,想象一条垂直的线从左至右扫描所有的建筑,根据图例我们不难发现,扫描到建筑的左边时,也就是有新的建筑出现,如果这个建筑的高度等于当前扫描的所有建筑的最高高度时,当前的这个坐标和高度应该加入到结果中,如果扫描到建筑的右边时,也就是有一个建筑离开了,此时扫描的建筑的最高高度发生了改变,则要把离开后的最高高度和当前坐标加入到结果中。

以上图为例,当扫描到红色建筑[3,15]时,此时我们应该保存扫描的高度集合,最大值恰好是15也就是红色建筑的高度,我们把[3,15]加入到天际线中。

当扫描到x=7时,也就是红色建筑[7,15]离开时,最高的高度由15变成了12,所以我们把[7,12]加入到天际线中。

那么为了模拟扫描建筑,实际上我们将表示建筑的线进行排序,然后再依次遍历即可。同时这里还有一个小技巧,例如红色建筑的两条线分别是[3, 15]和[7,15](x坐标,高度)我们把离开扫描线的建筑的边的高度记为负值,用来区分是进入还是离去,也就是[3, 15]和[7,-15],同时还要对所有的建筑按x的值进行升序排序,不过对于x相等的情况,我们来看一下特例:

也就是前一个建筑的右边和后一个建筑的左边在同一条线上,那么边的排序应该是如上图所示,也就是当x相同时,比较高度,高度大的排在前面。如果这里想不太清楚的同学,可以根据边的排序将两个建筑稍微重合一点或者分离一点,就能清楚的发现,高度长的边应该排在前面,只有这样才能正确求得天际线。

此外还有几种特殊情况我们来看一下:

这两种情况则是建筑左边横坐标相同,或者右边横坐标相同,此时排序的规则依旧是横坐标相同,按照高度的值!!降序排序,因为我们将离开的边编码为负数,所以在扫描到建筑离开的时候我们希望建筑高度低的排在前面,那么反映到编码上就是高度的值排序即可。以左边为例,如果将6,10排到6,12前面,显然不对,因为根据我们的规则,6,10也成为了天际线,但实际上只需要记录6,12即可。右边的图同理。

在存储高度时,我们使用树结构存储,可以保证删除操作在O(logn)内完成。

小技巧:存储高度集合中先加入0,当扫描建筑后没有建筑时,可以直接得到答案,而不用进行特殊处理。

程序:

C++

class Solution {
public:
    vector<vector<int>> getSkyline(vector<vector<int>>& buildings) {
        vector<pair<int, int>> lines;
        vector<vector<int>> res;
        multiset<int, greater<int>> h;
        for(auto &a:buildings){
            lines.push_back({a[0], a[2]});
            lines.push_back({a[1], -a[2]});
        }
        sort(lines.begin(), lines.end(), cmp());
        h.insert(0);
        for(auto iter = lines.begin(); iter != lines.end(); ++iter){
            int x = (*iter).first;
            int height = (*iter).second;
            int maxHeight = *h.begin();
            if(height > 0){
                if(height > maxHeight){
                    res.push_back({x, height});
                }
                h.insert(height);
            }else{
                height = -height;
                h.erase(h.find(height));
                if(maxHeight != *h.begin()){
                    res.push_back({x, *h.begin()});
                }
            }
        }
        return res;
    }
    struct cmp {
        bool operator() (const pair<int, int>& A, const pair<int, int>& B) {
            if (A.first < B.first) {
                return true;
            } else if (A.first == B.first) {
                return A.second > B.second;
            } else {
                return false;
            }
        }
    };
};

Java

import java.util.*;
class Solution {
    public List<List<Integer>> getSkyline(int[][] buildings) {
        List<Pair<Integer, Integer>> lines = new ArrayList<>();
        TreeMap<Integer, Integer> h = new TreeMap<>(new Comparator<Integer>() {
            @Override
            public int compare(Integer o1, Integer o2) {
                return o2 - o1;
            }
        });
        List<List<Integer>> res = new ArrayList<>();
        for(int[] a:buildings){
            lines.add(new Pair(a[0], a[2]));
            lines.add(new Pair(a[1], -a[2]));
        }
        Collections.sort(lines, new Comparator<Pair<Integer, Integer>>() {
            @Override
            public int compare(Pair<Integer, Integer> o1, Pair<Integer, Integer> o2) {
                int x1 = o1.getKey();
                int y1 = o1.getValue();
                int x2 = o2.getKey();
                int y2 = o2.getValue();
                if(x1 != x2)
                    return x1 - x2;
                else{
                    return y2 - y1;
                }
            }
        });
        h.put(0, 1);
        for(Pair p:lines){
            int x = (int)p.getKey();
            int height = (int)p.getValue();
            int maxHeight = h.firstKey();
            if(height > 0){
                if(height > maxHeight){
                    res.add(new ArrayList<>(Arrays.asList(x, height)));
                }
                h.put(height, h.getOrDefault(height, 0)+1);
            }else{
                height = -height;
                Integer v = h.get(height);
                if(v == 1){
                    h.remove(height);
                }else{
                    h.put(height, v-1);
                }
                if(maxHeight != h.firstKey()){
                    res.add(new ArrayList<>(Arrays.asList(x, h.firstKey())));
                }
            }
        }
        return res;
    }
}

 

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