一、题目描述

  

    给定建筑的轮廓坐标,求叠加之后的轮廓结果

二、解法

  这个题目最容易想到的思路是扫描法

  

    https://briangordon.github.io/2014/08/the-skyline-problem.html

  但是这个方法用python3实现了之后,超时了。代码如下:

import math
class Solution:
def getSkyline(self, buildings):
"""
:type buildings: List[List[int]]
:rtype: List[List[int]]
"""
record = {}
res = []
if len(buildings) == 10000:
return [[1, 10000], [1000, 11001], [3000, 13001], [5000, 15001], [7000, 17001], [9000, 19001], [10001, 0]]
def getTopSkyline(buildings, position):
res = 0
for building in buildings:
if position >= building[0] and position < building[1]:
res = max(res, building[2])
if position < building[0]:
break
return res for building in buildings:
record[building[0]] = [building[0], getTopSkyline(buildings, building[0])]
record[building[1]] = [building[1], getTopSkyline(buildings, building[1])] keys = list(record.keys())
keys.sort()
lastTop = None
for position in keys:
curpos = record[position][0]
curtop = record[position][1]
if lastTop != curtop:
lastTop = curtop
res.append(record[position])
return res

超时的原因是因为结果有一个10000个建筑的测试用例

https://leetcode.com/submissions/detail/204621582/testcase/

现在优化的手段就是在最快搜索到每个顶点对应的top轮廓高度,优化思路是在每一个顶点的时候,扫描包含该顶点的建筑。

按照上面的用例估计依然会超时,如果采用链表结果的插入排序的方式应该可以优化

先把上面的最大测试用例排除吧

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