An interesting problem! But not very easy at first glance. You need to think very clear about how will a keypoint be generated. Since I only learn from others' solutions and am still unable to give my personal explanations, I would suggest you to these solutions: C++ priority_queue solution, Python heapq solution. Of course, there is still a divide-and-conquer solution on Geeksforgeeks.com, which is much more difficult to understand :-)
Well, I just rewrite the code in the first solution as follows, by giving those variables more meaning names to help understand it.
1 class Solution { 2 public: 3 vector<pair<int, int>> getSkyline(vector<vector<int>>& buildings) { 4 int idx = 0, start, height, n = buildings.size(); 5 vector<pair<int, int> > skyline; 6 priority_queue<pair<int, int> > liveBuildings; 7 while (idx < n || !liveBuildings.empty()) { 8 if (liveBuildings.empty() || idx < n && buildings[idx][0] <= liveBuildings.top().second) { 9 start = buildings[idx][0]; 10 while (idx < n && buildings[idx][0] == start) { 11 liveBuildings.push(make_pair(buildings[idx][2], buildings[idx][1])); 12 idx++; 13 } 14 } 15 else { 16 start = liveBuildings.top().second; 17 while (!liveBuildings.empty() && liveBuildings.top().second <= start) 18 liveBuildings.pop(); 19 } 20 height = liveBuildings.empty() ? 0 : liveBuildings.top().first; 21 if (skyline.empty() || skyline.back().second != height) 22 skyline.push_back(make_pair(start, height)); 23 } 24 return skyline; 25 } 26 };
The Python translation is as follows. Note that since the default heapq is a min heap, we take the negative of the left and right positions and the height of buildings.
1 class Solution: 2 # @param {integer[][]} buildings 3 # @return {integer[][]} 4 def getSkyline(self, buildings): 5 idx, n = 0, len(buildings) 6 liveBuildings, skyline = [], [] 7 while idx < n or len(liveBuildings) > 0: 8 if len(liveBuildings) == 0 or (idx < n and buildings[idx][0] <= -liveBuildings[0][1]): 9 start = buildings[idx][0] 10 while idx < n and buildings[idx][0] == start: 11 heapq.heappush(liveBuildings, [-buildings[idx][2], -buildings[idx][1]]) 12 idx += 1 13 else: 14 start = -liveBuildings[0][1] 15 while len(liveBuildings) > 0 and -liveBuildings[0][1] <= start: 16 heapq.heappop(liveBuildings) 17 height = len(liveBuildings) and -liveBuildings[0][0] 18 if len(skyline) == 0 or skyline[-1][1] != height: 19 skyline.append([start, height]) 20 return skyline
