Course Schedule III 解法

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第 14 周题目
难度：Hard
LeetCode题号：630

题目

Description：

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.

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思考

维持变量totol用来保存学习了的课程总时间，course_selected保存选择了的课程对应的时间，最后返回course_selected的大小，即为所选的课程数目
由题意可知，我们应该每次选择d最小的那行数据（假设叫current），判断这个数据加上之前的那些课程的时间totol是否超过了这行数据的d，如果没超过，就选这门课程，更新course_selected和totol
如果超过了这行数据的d，那么我们就要考虑上一个数据（假设为pre），比较他们的t，选择较小的课。因为current的d比较大，所以是可以的

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代码

bool comp(const vector<int> & a, const vector<int> & b) {    return a[1] < b[1];}class Solution {public:    int scheduleCourse(vector<vector<int>>& courses) {        sort(courses.begin(), courses.end(), comp);        int total_day = 0;        multiset<int> course_selected;        for (int i = 0; i < courses.size(); i++) {            if (total_day + courses[i][0] <= courses[i][1]) {                total_day += courses[i][0];                course_selected.insert(courses[i][0]);            } else if (*course_selected.rbegin() > courses[i][0]) {                total_day += courses[i][0] - *course_selected.rbegin();                course_selected.erase(--course_selected.end());                course_selected.insert(courses[i][0]);            }        }        return course_selected.size();    }};