[LeetCode 207] Course Schedule

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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;    }



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