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Friday, July 21, 2017

[Leetcode] Course Schedule III, Solution

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:
  1. The integer 1 <= d, t, n <= 10,000.
  2. You can't take two courses simultaneously.

[Thoughts]

这道题并不难。题目说的很清楚,主要是看course是否能在deadline前上完。对deadline进行排序,然后根据时间轴填时间段即可。

以题目中的case为例,

根据deadline排序完之后,结果是 [100, 200], [1000, 1250], [200, 1300], [2000, 3200]

首先起始时间从0开始,填第一个course,结束时间只是100,小于预期的200,所以这门课可以读。

填第二个course,时长1000,结束时间是1100,小于1250,所以这门课也可以读

填第三个course, 时长200, 结束时间是1300, 等于预期的1300, 这门课可读

填第四个course,时长2000, 结束时间是3300, 大于预期的3200,这门课不可读

所以,最终只有三门课可读。


[Code]
1:    int scheduleCourse(vector<vector<int>>& courses) {  
2:      sort(courses.begin(), courses.end(), [](vector<int>& a, vector<int>& b){return a[1]< b[1];} );  
3:    
4:      int now = 0;  
5:      priority_queue<int> used;  
6:      for(int i =0; i< courses.size(); i++) {  
7:        used.push(courses[i][0]);  
8:        now += courses[i][0];  
9:        if(now > courses[i][1]) {  
10:          now -= used.top();  
11:          used.pop();  
12:        }  
13:      }  
14:      return used.size();  
15:    }  


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