[LeetCode] Course Schedule III 课程清单之三
There are n
different online courses numbered from 1
to n
. Each course has some duration(course length) t
and closed on dth
day. A course should be taken continuously for t
days and must be finished before or on the dth
day. You will start at the 1st
day.
Given n
online courses represented by pairs (t,d)
, your task is to find the maximal number of courses that can be taken.
Example:
Input: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
Output: 3
Explanation:
There're totally 4 courses, but you can take 3 courses at most:
First, take the 1st course, it costs 100 days so you will finish it on the 100th day, and ready to take the next course on the 101st day.
Second, take the 3rd course, it costs 1000 days so you will finish it on the 1100th day, and ready to take the next course on the 1101st day.
Third, take the 2nd course, it costs 200 days so you will finish it on the 1300th day.
The 4th course cannot be taken now, since you will finish it on the 3300th day, which exceeds the closed date.
Note:
- The integer 1 <= d, t, n <= 10,000.
- You can't take two courses simultaneously.
这道题给了我们许多课程,每个课程有两个参数,第一个是课程的持续时间,第二个是课程的最晚结束日期,让我们求最多能上多少门课。博主尝试了递归的暴力破解,TLE了。这道题给的提示是用贪婪算法,那么我们首先给课程排个序,按照结束时间的顺序来排序,我们维护一个当前的时间,初始化为0,再建立一个优先数组,然后我们遍历每个课程,对于每一个遍历到的课程,当前时间加上该课程的持续时间,然后将该持续时间放入优先数组中,然后我们判断如果当前时间大于课程的结束时间,说明这门课程无法被完成,我们并不是直接减去当前课程的持续时间,而是取出优先数组的顶元素,即用时最长的一门课,这也make sense,因为我们的目标是尽可能的多上课,既然非要去掉一门课,那肯定是去掉耗时最长的课,这样省下来的时间说不定能多上几门课呢,最后返回优先队列中元素的个数就是能完成的课程总数啦,参见代码如下:
class Solution {
public:
int scheduleCourse(vector<vector<int>>& courses) {
int curTime = ;
priority_queue<int> q;
sort(courses.begin(), courses.end(), [](vector<int>& a, vector<int>& b) {return a[] < b[];});
for (auto course : courses) {
curTime += course[];
q.push(course[]);
if (curTime > course[]) {
curTime -= q.top(); q.pop();
}
}
return q.size();
}
};
类似题目:
参考资料:
https://discuss.leetcode.com/topic/93790/short-java-code-using-priorityqueue
https://discuss.leetcode.com/topic/93712/python-straightforward-with-explanation
https://discuss.leetcode.com/topic/93884/c-short-elegant-o-nlogn-time-o-k-space-solution/2
LeetCode All in One 题目讲解汇总(持续更新中...)
[LeetCode] Course Schedule III 课程清单之三的更多相关文章
- [LeetCode] Course Schedule II 课程清单之二
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prer ...
- [LeetCode] 210. Course Schedule II 课程清单之二
There are a total of n courses you have to take, labeled from 0 to n-1. Some courses may have prereq ...
- [LeetCode] Strobogrammatic Number III 对称数之三
A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside ...
- [LeetCode] Combination Sum III 组合之和之三
Find all possible combinations of k numbers that add up to a number n, given that only numbers from ...
- [LeetCode] Basic Calculator III 基本计算器之三
Implement a basic calculator to evaluate a simple expression string. The expression string may conta ...
- [LeetCode] Course Schedule 课程清单
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prer ...
- [LeetCode] 207. Course Schedule 课程清单
There are a total of n courses you have to take, labeled from 0 to n-1. Some courses may have prereq ...
- LeetCode Course Schedule II
原题链接在这里:https://leetcode.com/problems/course-schedule-ii/ 题目: There are a total of n courses you hav ...
- [LeetCode] 210. Course Schedule II 课程安排II
There are a total of n courses you have to take, labeled from 0 to n - 1. Some courses may have prer ...
随机推荐
- Java实现单向链表基本功能
一.前言 最近在回顾数据结构与算法,有部分的算法题用到了栈的思想,说起栈又不得不说链表了.数组和链表都是线性存储结构的基础,栈和队列都是线性存储结构的应用- 本文主要讲解单链表的基础知识点,做一个简单 ...
- 痛吻过YY寻找到真爱的三非渣本春招之路
写下这篇文章可能就不是大家乐于见闻的面经了,更多是深入一些面试细节. 前言 我猜拿到了BAT等一线互联网公司Offer的小伙伴或者那些老鸟看到这条标题的时候会不屑一顾,认为YY这种级别的公司是属于二线 ...
- 使用SQLiteOpenHelper类对数据库简单操作
实现数据库基本操作 数据库创建的问题解决了,接下来就该使用数据库实现应用程序功能的时候了.基本的操作包括创建.读取.更新.删除,即我们通常说的CRUD(Create, Read, Upda ...
- 爬取博主所有文章并保存到本地(.txt版)--python3.6
闲话: 一位前辈告诉我大学期间要好好维护自己的博客,在博客园发布很好,但是自己最好也保留一个备份. 正好最近在学习python,刚刚从py2转到py3,还有点不是很习惯,正想着多练习,于是萌生了这个想 ...
- c语言程序设计第6周编程作业一(分解质因数)
分解质因数 题目内容: 每个非素数(合数)都可以写成几个素数(也可称为质数)相乘的形式,这几个素数就都叫做这个合数的质因数.比如,6可以被分解为2x3,而24可以被分解为2x2x2x3. 现在,你的程 ...
- C语言程序设计(基础)最后一次作业-- 总结报告
本次作业是本学期的最后一次作业,有始有终,本次作业回顾下本学期的第0次作业,回答下面几个问题. 注意:在回答问题时请不要简单的回答 "不是","是".请把这当成 ...
- http post/get 2种使用方式
public class HttpUtil { //HttpPost public static String executePost(String url, List<NameValue ...
- 翻译:CREATE FUNCTION语句(已提交到MariaDB官方手册)
本文为mariadb官方手册:CREATE FUNCTION的译文. 原文:https://mariadb.com/kb/en/library/create-function/我提交到MariaDB官 ...
- JAVA_SE基础——30.构造代码块
黑马程序员入学blog...构造代码块作用:给所有的对象进行统一的初始化. 问题:要求每个小孩出生都会哭,这份代码有两个构造函数,如果需要每个小孩出生都要哭的话,那么就需要在不同的构造函数中都调用cr ...
- ubuntu启动报/root/.profile mesg:ttyname failed错误的解决办法
修改/root/.profile文件,如下命令 sudo gedit /root/profile 将文中的最后一行mesg n修改成tty -s && mesg n