Course Schedule II

Topic:

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, return the ordering of courses you should take to fini SH all courses.

There may is multiple correct orders, you just need to return one of the them. If It is impossible-to-finish all courses, return an empty array.

For example:

2, [[1,0]]

There is a total of 2 courses to take. To take course 1 should has finished course 0. So the correct course order is[0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There is a total of 4 courses to take. To take course 3 should has finished both courses 1 and 2. Both Courses 1 and 2 should is taken after you finished course 0. So one correct course order is [0,1,2,3] . Another correct ordering is [0,2,1,3] .

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more on how a graph is represented.

Click to show more hints.

Hints:

1. This problem was equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it'll be impossible to take all courses.
2. Topological Sort via dfs-a great video tutorial (minutes) on Coursera explaining the basic concepts of topological so Rt.
3. Topological sort could also is done via BFS.

Links: http://leetcode.com/problems/course-schedule-ii/

Exercises

Like course schedule, this is the order in which topological sorting is obtained. We are still using Kahn ' s algorithm and Tarjan's algorithm.

Kahn ' s algorithm:

 Public classSolution { Public int[] FindOrder (intNumcourses,int[] Prerequisites) {//Kahn ' s algorithms        if(numcourses <= 0)            return New int[]{}; int[] res =New int[numcourses];  for(inti = 0; i < numcourses; i++) Res[i]=i; if(Prerequisites = =NULL|| Prerequisites.length = = 0)            returnRes; int[] Indegree =New int[numcourses];  for(int[] edge:prerequisites) indegree[edge[1]]++; Queue<Integer> queue =NewLinkedlist<>();  for(inti = 0; i < numcourses; i++)//Calculate Indegree            if(Indegree[i] = = 0) Queue.offer (i); intindex = numCourses-1;  while(!Queue.isempty ()) {            intSource =Queue.poll (); Res[index--] = source;//Reverse Post Order             for(int[] edge:prerequisites) {                if(Edge[0] = =source) {indegree[edge[1]]--; if(Indegree[edge[1]] = = 0) Queue.offer (edge[1]); }            }        }                if(Index < 0) {//looped through all vertex            returnRes; } Else            return New int[]{};//Have cycle, not DAG    }}

Reference:

Https://leetcode.com/discuss/49696/easy-understanding-dfs-solution-in-c

Https://leetcode.com/discuss/36572/java-code-for-course-schedule-ii

Https://leetcode.com/discuss/42710/java-dfs-double-cache-visiting-each-vertex-once-433ms

Https://leetcode.com/discuss/35605/two-ac-solution-in-java-using-bfs-and-dfs-with-explanation

Https://leetcode.com/discuss/35787/simple-ac-c-dfs-toposort-solution

Https://leetcode.com/discuss/35590/dfs-solution-in-java

Course Schedule II