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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:
[0, 10000]
.Li
.[...[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], ...]
Credits:
public class Solution { public List<int[]> getSkyline(int[][] buildings) { /* 这题有个很巧的思路:离散化。 什么意思呢?既然每栋大楼的高和左右边界都是整数,那么不妨把线段用一个个整点表示。既然最后只求一个轮廓,那么对每个横坐标,就记录纵坐标 (高度)最大的点。最后从左到右扫描一遍,当高度发生变化时就输出。 但是[[0,2147483647,2147483647]]结果会TLE */ Map<Integer,Integer> map=new HashMap<Integer,Integer>(); List<int[]> res=new ArrayList<int []>(); if(buildings==null||buildings.length<=0) return res; //int h[]=new int[buildings[buildings.length-1][1]]; for(int i=0;i<buildings.length;i++){ for(int j=buildings[i][0];j<buildings[i][1];j++){ if(!map.isEmpty()&&map.containsKey(j)){ int h=map.get(j); h=Math.max(buildings[i][2],h); map.put(j,h); }else{ map.put(j,buildings[i][2]); } } } for(int i=1;i<map.size();i++){ if(map.get(i)!=map.get(i-1)){ res.add(new int[]{i,map.get(i)}); } } return res; } }
[leedcode 218] The Skyline Problem
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原文地址:http://www.cnblogs.com/qiaomu/p/4709434.html