【LeetCode从零单排】No 630.Course Schedule III

There are n different online courses numbered from 1 to n. Each course has some duration(course length) t and closed on dthday. 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.

思路：

    贪心算法，将所给的课程按照结束的时间进行升序排序。

     依次取出排序后的课程，如果当前的时间（从第0天开始）加上该课程的上课的天数不大于课程结课的时间，即该门课可以被上完，所以选取该门课，否则去掉该门课。

代码：

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