Course Schedule--Leetcode

There is a total of n courses you have to take, labeled from 0 to n - 1 .

Some courses May has prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a Pair[0,1]

Given the total number of courses and a list of prerequisite pairs, are it possible for your to finish all courses?

For example:

2, [[1,0]]

There is a total of 2 courses to take. To take course 1 should has finished course 0. So it is possible.

2, [[1,0],[0,1]]

There is a total of 2 courses to take. To take course 1 should has finished course 0, and to take course 0 you should also has finished course 1. So it is impossible.

Basic idea: Topology sequencing

There are dependencies between courses.

1. Find out what pre-conditions are available for learning.

2. After completing this course, update other courses that depend on the course.

3. Repeat steps 1 and 2.

In the algorithm, a degree array is maintained, or an array of pre-degree classes is called. Indicates the number of pre-courses that need to be studied in order to learn the course.

If the value is 0, the course can start learning.

Class Solution {public:    bool Canfinish (int numcourses, vector<pair<int, int>>& prerequisites) {        vector<int> degree (numcourses);        Unordered_set<int> finished_courses;        for (auto p:prerequisites)            ++degree[p.first];                    for (int i=0; i<numcourses; i++) {            if (!degree[i])                Finished_courses.insert (i);        }                int count = 0;        while (!finished_courses.empty ()) {            int course = *finished_courses.begin ();            Finished_courses.erase (Finished_courses.begin ());            for (auto p:prerequisites) {                if (P.second = = Course &&!--degree[p.first])                    Finished_courses.insert ( P.first);            }            ++count;        }                return count = = numcourses;    }};

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Course Schedule--Leetcode