LWC 54:699. Falling Squares

LWC 54:699. Falling Squares

传送门:699. Falling Squares

Problem:

On an infinite number line (x-axis), we drop given squares in the order they are given. The i-th square dropped (positions[i] = (left, side_length)) is a square with the left-most point being positions[i][0] and sidelength positionsi. The square is dropped with the bottom edge parallel to the number line, and from a higher height than all currently landed squares. We wait for each square to stick before dropping the next. The squares are infinitely sticky on their bottom edge, and will remain fixed to any positive length surface they touch (either the number line or another square). Squares dropped adjacent to each other will not stick together prematurely. Return a list ans of heights. Each height ans[i] represents the current highest height of any square we have dropped, after dropping squares represented by positions[0], positions1, …, positions[i].

Example 1:

Input: [[1, 2], [2, 3], [6, 1]] Output: [2, 5, 5] Explanation: After the first drop of positions[0] = [1, 2]: _aa _aa The maximum height of any square is 2. After the second drop of positions1 = [2, 3]: __aaa __aaa __aaa aa_ aa_ The maximum height of any square is 5. The larger square stays on top of the smaller square despite where its center of gravity is, because squares are infinitely sticky on their bottom edge. After the third drop of positions1 = [6, 1]: __aaa __aaa __aaa _aa _aa___a The maximum height of any square is still 5. Thus, we return an answer of [2, 5, 5].

Example 2:

Input: [[100, 100], [200, 100]] Output: [100, 100] Explanation: Adjacent squares don’t get stuck prematurely - only their bottom edge can stick to surfaces.

Note:

1 <= positions.length <= 1000.

1 <= positions0 <= 10^8.

1 <= positions1 <= 10^6.

思路: 有点几何题的味道,实际上正方形往上叠总共就5种情况,如下:

前四种情况,可以合并成一种情况处理,只需要搜索蓝色两条边界内是否存在对应的边,如果有,则说明需要叠加。第五种情况需要特殊处理,直接扫描所有正方形,包含蓝色矩形块则更新高度。

代码如下:

    class Edge{
        int x;
        int y;
        int h;

        Edge(int x, int y, int h){
            this.x = x;
            this.y = y;
            this.h = h;
        }
    }

    public List<Integer> fallingSquares(int[][] positions) {
        int n = positions.length;
        List<Integer> ans = new ArrayList<>();

        TreeMap<Double, Integer> map = new TreeMap<>();
        List<Edge> heights = new ArrayList<Edge>();
        int max = 0;

        for (int i = 0; i < n; ++i) {
            int x = positions[i][0];
            int h = positions[i][1];
            int y = h + x - 1;


            int h_max = 0;
            for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) {
                h_max = Math.max(h_max, map.get(e));
            }

            for (Edge edge : heights) {
                if (edge.x <= x && edge.y >= y) {
                    h_max = Math.max(h_max, edge.h);
                }
            }

            h_max += h;

            map.put(x * 1.0, h_max);
            map.put(y * 1.0, h_max);
            heights.add(new Edge(x, y, h_max));

            max = Math.max(max, h_max);
            ans.add(max);
        }

        return ans;
    }       

其实吧,上述五种情况也能合并在一块,比较取巧,先把对应的所有边都枚举出来,之后不断在其基础上累加高度即可。造房子的感觉!

代码如下:

        public List<Integer> fallingSquares(int[][] positions) {
            int n = positions.length;
            List<Integer> ans = new ArrayList<>();

            TreeMap<Double, Integer> map = new TreeMap<>();
            for (int[] pos : positions) {
                int x = pos[0];
                int h = pos[1];
                int y = x + h - 1;
                map.put(x * 1.0, 0);
                map.put(y * 1.0, 0);
            }
            int max = 0;

            for (int i = 0; i < n; ++i) {
                int x = positions[i][0];
                int h = positions[i][1];
                int y = h + x - 1;


                int h_max = 0;
                for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) {
                    h_max = Math.max(h_max, map.get(e));
                }

                h_max += h;

                for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) {
                    map.put(e, h_max);
                }

                max = Math.max(max, h_max);
                ans.add(max);
            }

            return ans;
        }   

本文参与腾讯云自媒体分享计划,欢迎正在阅读的你也加入,一起分享。

发表于

我来说两句

0 条评论
登录 后参与评论

相关文章

来自专栏别先生

Redis的搭建和Redis的集群搭建

1、Redis的官网:https://redis.io/     Redis的测试网站:http://try.redis.io/

1110
来自专栏小工匠技术圈

【Java小工匠聊密码学】--数字签名-DSA

DSA(Digital Signature Algorithm)是Schnorr和ElGamal签名算法的变种,被美国NIST作为数字签名标准(DigitalS...

712
来自专栏pangguoming

API手册 常用功能

directive [ng] a form input input [checkbox] input [email] input [number] input ...

3179
来自专栏别先生

INFO JobScheduler: Added jobs for time 1524468752000 ms/INFO MemoryStore: Block input-0-152446914300

34312
来自专栏Hongten

python开发_tkinter_窗口控件_自己制作的Python IDEL_博主推荐

在了解python中的tkinter模块的时候,你需要了解一些tkinter的相关知识

882
来自专栏大数据学习笔记

org.springframework.beans.factory.NoSuchBeanDefinitionException: No bean named 'userService' availab

1.问题描述 Hibernate+Spring+SpirngMVC+Maven异常信息: org.springframework.beans.factory...

4469
来自专栏搞前端的李蚊子

Html5模拟通讯录人员排序(sen.js)

// JavaScript Document  var PY_Json_Str = ""; var PY_Str_1 = ""; var PY_Str_...

5256
来自专栏互联网研发闲思录

java 使用CRF遇到的问题汇总

1、libCRFPP.so放在idea项目 resources下,打jar包时打在jar中。        jar包工具类 /* * Class Native...

20010
来自专栏别先生

Redis的搭建和Redis的集群搭建

1、Redis的官网:https://redis.io/     Redis的测试网站:http://try.redis.io/

1181
来自专栏互联网软件技术

省市区联动

923

扫码关注云+社区