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Golang语言社区

问题帖子--Concurrent Read/Write Map

DK1.5 引入了 concurrent package, 提供了更多的concurrent 控制方法。 还提供了一个 ConcurrentHashMap 类...

36412
来自专栏编程

Python编程语言入门经典案例

【程序1】 题目:输入一行字符,分别统计出其中英文字母、空格、数字和其它字符的个数。 1#!/usr/bin/python 2#-*- coding:utf-8...

3240
来自专栏机器学习实践二三事

leetcode之-题19

题目 [图片] Given a linked list, remove the nth node from the end of list and re...

1977
来自专栏糊一笑

Immutable日常操作之深入API

写在前面 本文只是个人在熟悉Immutable.js的一些个人笔记,因此我只根据我自己的情况来熟悉API,所以很多API并没有被列举到,比如常规的push/ma...

4029
来自专栏DT乱“码”

正则判断工具类

package com.gulf.utils; import java.text.ParseException; import java.text.Simpl...

23210
来自专栏老马说编程

(41) 剖析HashSet / 计算机程序的思维逻辑

查看历史文章,请点击上方链接关注公众号。 上节介绍了HashMap,提到了Set接口,Map接口的两个方法keySet和entrySet返回的都是Set,本节,...

1869
来自专栏Java3y

TreeMap就这么简单【源码剖析】

2155
来自专栏小筱月

JavaScript 数组去重

解决方案有很多,可以是两个 for 循环,或者一个 for 和 一个 filter,一个 filter 和 一个 every,接下来介绍这几种方法:

1656
来自专栏Bingo的深度学习杂货店

Q101 Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric aroun...

3268
来自专栏菩提树下的杨过

XStream、JAXB 日期(Date)、数字(Number)格式化输出xml

XStream、Jaxb是java中用于对象xml序列化/反序列化 的经典开源项目,利用它们将对象转换成xml时,经常会遇到日期(Date)、数字按指定格式输出...

2577

扫码关注云+社区