[Leedcode 210] Course Schedule II

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] .

 Public classSolution { Public int[] FindOrder (intNumcourses,int[] Prerequisites) {     //The subject is the topology map, verifying that the graph has a ring, starting from the leaf node (the point of 0), if all the points can be traversed, the condition is satisfied//An auxiliary array is required, the subscript of the array represents the course number, and the value of the array represents the degree//The queue holds a point with a degree of 0, and each time a count is added to the inside, the result is saved to the list each time it pops out of the queue. Final reverse output result        int[]flag=New int[numcourses]; int[] res=New int[numcourses]; List<Integer> list=NewArraylist<integer>();  for(inti=0;i<prerequisites.length;i++) {flag[prerequisites[i][1]]++; } LinkedList<Integer> queue=NewLinkedlist<integer>(); intCount=0;  for(inti=0;i<numcourses;i++){            if(flag[i]==0) {//the degree of 0Queue.add (i); Count++; }        }         while(!Queue.isempty ()) {            intk=Queue.remove ();            List.add (k);  for(inti=0;i<prerequisites.length;i++){                if(k==prerequisites[i][0]){                    intL=prerequisites[i][1]; FLAG[L]--; if(flag[l]==0) {Count++;                    Queue.add (l); }                }                                }                                }         for(intI=0;i<list.size (); i++) {Res[i]=list.get (List.size ()-1-i); }        if(count==numcourses)returnRes; Else return New int[0]; }}

[Leedcode 210] Course Schedule II