[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) 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), 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 = 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();
    }
};