refactor 210

Author: fishercoder1534

src/main/java/com/fishercoder/solutions/_210.java

@@ -6,41 +6,6 @@
 import java.util.Queue;
 import java.util.Set;
 
-/**
- * 210. Course Schedule II
- *
- * There are a total of n courses you have to take, labeled from 0 to n - 1.
- Some courses may have 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 finish all courses.
-
- There may be multiple correct orders,
- you just need to return one of them.
- If it is impossible to finish all courses, return an empty array.
-
- For example:
-
- 2, [[1,0]]
- There are a total of 2 courses to take.
- To take course 1 you should have finished course 0. So the correct course order is [0,1]
-
- 4, [[1,0],[2,0],[3,1],[3,2]]
- There are a total of 4 courses to take.
- To take course 3 you should have finished both courses 1 and 2.
- Both courses 1 and 2 should be 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 about how a graph is represented.
- You may assume that there are no duplicate edges in the input prerequisites.
- click to show more hints.
-
- Hints:
- This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
- Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
- Topological sort could also be done via BFS.
- */
 public class _210 {
 
     public static class Solution1 {