[LeetCode 207] Course Schedule

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, is it possible for you to finish all courses?

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 it is possible.

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

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

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

topological sort, in a directed graph, every time found node without pre-node, then remove all its edges. until iterate all nodes or found cycle.

public boolean canFinish(int numCourses, int[][] prerequisites) {
        List<HashSet> preCourse = new ArrayList<HashSet>();
        for(int i=0;i<numCourses;i++){
            preCourse.add(new HashSet<Integer>());
        }
        for(int i=0;i<prerequisites.length;i++){
            preCourse.get(prerequisites[i][1]).add(prerequisites[i][0]);
        }
        int[] prenums = new int[numCourses];
        for(int i=0;i<numCourses;i++){
            Set set = preCourse.get(i);
            Iterator<Integer> iterator = set.iterator();
            while(iterator.hasNext()){
                prenums[iterator.next()]++;
            }
        }
        for(int i=0;i<numCourses;i++){
            int j=0;
            for(;j<numCourses;j++){
                if(prenums[j]==0) break;
            }
            if(j == numCourses) return false;
            prenums[j] = -1;
            Set set = preCourse.get(j);
            Iterator<Integer> iterator = set.iterator();
            while(iterator.hasNext()){
                prenums[iterator.next()]--;
            }
        }
        return true;
    }