[Array]218. The Skyline Problem

2019-01-31  本文已影响0人  野生小熊猫

218. The Skyline Problem

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).

image image

The geometric information of each building is represented by a triplet of integers [Li, Ri, Hi], where Liand 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:

代码:

方法:

class Solution:
    def getSkyline(self, buildings):
        """
        :type buildings: List[List[int]]
        :rtype: List[List[int]]
        """
        ans=[]
        if buildings==None or len(buildings)==0:
            return ans
        
        events=[]
        events1=[]
        for b in buildings:
            #加入entering事件
            events1.append([b[0],b[2],"e"])
        events1.sort(key=lambda x:x[1],reverse=True)
            #加入leaving事件
        events2=[]
        for b in buildings:
            events2.append([b[1],b[2],"l"])
        events2.sort(key=lambda x:x[1])
        #给事件排序
        #先entering再leaving
        events=events1+events2
        events.sort(key=lambda x:x[0])
        
        heights=[0]
        for e in events:
            if e[2]=='e':
                #进入事件
                if e[1]>heights[-1]:
                    ans.append([e[0],e[1]])
                heights.append(e[1])
                heights.sort()
            else:
                #出去事件
                heights.remove(e[1])
                if e[1]>heights[-1]:
                    ans.append([e[0],heights[-1]])
            
        return ans

讨论:

1.把节点找出来非常重要
2.reference讲解:花花酱

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