There are n different online courses numbered from 1 to n. Each course has some duration(course length) tand 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)

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.

這道題給了我們許多課程，每個課程有兩個引數，第一個是課程的持續時間，第二個是課程的最晚結束日期，讓我們求最多能上多少門課。博主嘗試了遞迴的暴力破解，TLE了。這道題給的提示是用貪婪演算法，那麼我們首先給課程排個序，按照結束時間的順序來排序，我們維護一個當前的時間，初始化為0，再建立一個優先陣列，然後我們遍歷每個課程，對於每一個遍歷到的課程，當前時間加上該課程的持續時間，然後將該持續時間放入優先陣列中，然後我們判斷如果當前時間大於課程的結束時間，說明這門課程無法被完成，我們並不是直接減去當前課程的持續時間，而是取出優先陣列的頂元素，即用時最長的一門課，這也make sense，因為我們的目標是儘可能的多上課，既然非要去掉一門課，那肯定是去掉耗時最長的課，這樣省下來的時間說不定能多上幾門課呢，最後返回優先佇列中元素的個數就是能完成的課程總數啦，參見程式碼如下：

class Solution {
public:
    int scheduleCourse(vector<vector<int>>& courses) {
        int curTime = 0;
        priority_queue<int> q;
        sort(courses.begin(), courses.end(), [](vector<int>& a, vector<int>& b) {return a[1] < b[1];});
        for (auto course : courses) {
            curTime += course[0];
            q.push(course[0]);
            if (curTime > course[1]) {
                curTime -= q.top(); q.pop();
            }
        }
        return q.size();
    }
};

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