LeetCode Course Schedule III

LeetCode 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 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.


Input: [[100, 200], [200, 1300], [1000, 1250], [2000, 3200]]
Output: 3
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.


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


使用贪心的思路, 首先根据经验,越早结束的课程,越早选修并结束该课程越好,因为这样可以留出更多的时间选修其他课程。所以我们首先对所有课程按关闭时间从小到大排序,每次我们贪心的选择最早关闭的课程。




  1. 选修第一门课程,now=3<8,优先队列堆顶为3
  2. 选修第二门课程,now=3+7=10<=10,优先队列堆顶为7
  3. 选修第三门课程,now=10+5=15>14,失败;发现堆顶课程的持续时间是7,大于当前课程的持续时间5,所以可以把堆顶课程换成该门课程,则新的now=10-7+5=8,堆顶为5。因为该门课程的持续时间比堆顶短,且关闭时间已排序,所以大于堆顶的关闭时间,所以把堆顶课程换成该课程,该课程肯定能过完成。且新的now=8比先前的now=10要小,使得可以完成第四门课程。
  4. 选修第四门课,now=8+8=16<17,优先队列为8。


class Solution {
	int scheduleCourse(vector<vector<int>>& courses) {
		auto cmp = [](vector<int>& c1, vector<int>& c2) {return c1[1] < c2[1]; };
		sort(courses.begin(), courses.end(), cmp);
		int now = 0, ans = 0;
		priority_queue<int> pq;
		for (const auto& c : courses) {
			if (now + c[0] <= c[1]) {
				now += c[0];
			else if (!pq.empty() && c[0] < pq.top()) {
				now = now - pq.top() + c[0];

		return ans;


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