# Android Room 科普介绍
在 Android 开发中，数据的存储和管理是一个至关重要的部分。Android Room 是一个来简化 SQLite 数据库交互的库，它提供了一个更为高效和易于使用的 API，能帮助开发者更方便地管理应用的数据。
## 什么是 Android Room？
Android Room 是 Google 在 Android Jetpack 中推出的一部分，
原创
2024-08-13 07:35:56
16阅读
Interval as an array: Previous interval as an object: Implement a Comparator<Interval> Syntax: don't forget the public sign when defining a function
转载
2015-12-22 08:32:00
98阅读
2评论
A and B and Lecture Rooms题意要求我们找有多少个点iii满足dis(i,x),dis(i,y)dis(i, x),
原创
2021-08-26 16:36:24
58阅读
Game RoomsTime Limit: 4000/4000MS (Java/Others) Memory Limit: 65535/65535KB (Java/Others)Your company has just constructed a new skyscraper, but you ...
转载
2015-11-03 20:03:00
79阅读
2评论
/** * Definition for an interval. * public class Interval { * int start; * int end; * Interval() { start = 0; end = 0; } * Interval(int s, int e) { st
转载
2018-07-18 12:45:00
73阅读
2评论
841. Keys and Rooms**https://leetcode.com/problems/keys-and-rooms/题目描述There are N rooms and you start in room 0. Each room has a distinct nu
原创
2022-05-30 10:22:57
188阅读
Like the previous one Meeting Room, still need to sort the intervals using a comparator. We need to simulate the process to know the maximum number of
转载
2015-12-22 09:19:00
102阅读
2评论
题目大意：有n个房间，n！个钥匙，在房间中，最多可以破k扇门，然后得到其中的钥匙，去开其它的门，但是第一扇门不可以破开，求可以打开所有门的概率。题解：首先，建立这样的一个模型，题目相当于给出一个图，求形成1--K个环的可能性有多大。但是节点1不可以形成子环。那么首先，n个点形成1--k个环就是第一类斯特灵数的定义，但是该如何处理1的问题呢，既然算起来比较麻烦，那么正难则反，减去节点1成为自环的情况就可以了。第一类斯特林公式：S（m,n）=(m-1)*S(m-1,n)+S(m-1,n-1)。#include #include using namespace std;long long ans[2
转载
2014-01-18 09:10:00
33阅读
题目There are N rooms and you start in room 0. Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room.Formally, each room
There are N rooms and you start in room 0. Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next r
转载
2021-03-21 02:44:00
73阅读
2评论
"Link" 先不管$1$号点的限制，$k$次破能够全部打开的充要条件是构成了$\le k$个环。而$1$号点无法被打开的充要条件是自成一环。 因此答案就是$\frac{\sum\limits_{i=1}^k\left[_i^n\right]\left[_{i 1}^{n 1}\right]}{n!
转载
2020-01-29 07:56:00
63阅读
2评论
题目：求从s開始的递增序列（每次加1）。求出他们加和不小于D的那个最后的加数。 分析：数学题。分治。s + s+1 + ... + n = n*（n+1）/2 - s*（s-1）/2 = （n+s）*（n-s+1）/2。 直接二分答案就可以（二分范围0~10^8）。 说明：(⊙_⊙)。 #inclu
转载
2017-04-25 11:04:00
609阅读
2评论
Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), determine if a person could attend all mee
转载
2019-10-31 23:31:00
212阅读
2评论
原题链接在这里：https://leetcode.com/problems/meeting-rooms/ 题目： Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],.
转载
2016-02-22 11:31:00
481阅读
2评论
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;const int maxn = 5000;const ll INF = 1e18;int T, n;ll date[2]
原创
2022-08-17 15:34:21
33阅读
if the new interval's start is later than the smallest end index in our pq. then we can use the same room, and update the end , put it back to the pq.
转载
2018-08-09 17:03:00
119阅读
2评论
原题链接在这里：https://leetcode.com/problems/keys-and-rooms/ 题目： There are N rooms and you start in room 0. Each room has a distinct number in 0, 1, 2, ...,
转载
2019-08-04 08:47:00
89阅读
Sort the vector by start time, then check if s[i].start < s[i + 1].end
原创
2023-08-23 09:13:46
64阅读
Give you set of meetings start time and end time, count how many meeting rooms needed. For example: We can use a memo table to store the at which time
转载
2019-03-27 02:03:00
85阅读
2评论
原题链接在这里：https://leetcode.com/problems/meeting-rooms-ii/ 题目： Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2
转载
2016-02-22 11:51:00
273阅读
2评论
