Parallel Courses

There are N courses, labelled from 1 to N.

We are given relations[i] = [X, Y], representing a prerequisite relationship between course X and course Y: course X has to be studied before course Y.

In one semester you can study any number of courses as long as you have studied all the prerequisites for the course you are studying.

Return the minimum number of semesters needed to study all courses.  If there is no way to study all the courses, return -1.

Example 1:

Input: N = 3, relations = [[1,3],[2,3]]
Output: 2
Explanation: 
In the first semester, courses 1 and 2 are studied. In the second semester, course 3 is studied.

思路：就是topo排序和level order，注意indegree要用array來表示就可以了；x -> y, x的neighbor加入y；y 的indegree ++；

class Solution {
    public int minimumSemesters(int N, int[][] relations) {
        if(N < 0 || relations == null) {
            return -1;
        }
        HashMap<Integer, List<Integer>> graph = new HashMap<>();
        for(int i = 1; i <= N; i++) {
            graph.putIfAbsent(i, new ArrayList<Integer>());
        }
        // indegree 用array表示即可；
        int[] indegree = new int[N + 1];
        for(int i = 0; i < relations.length; i++) {
            int x = relations[i][0];
            int y = relations[i][1];
            graph.get(x).add(y);
            indegree[y]++;
        }
        
        Queue<Integer> queue = new LinkedList<Integer>();
        for(int i = 1; i <= N; i++) {
            if(indegree[i] == 0) {
                queue.offer(i);
            }
        }
        
        int step = 0;
        int count = 0;
        while(!queue.isEmpty()) {
            int size = queue.size();
            for(int i = 0; i < size; i++) {
                count++;
                Integer node = queue.poll();
                for(Integer neighbor: graph.get(node)) {
                    indegree[neighbor]--;
                    if(indegree[neighbor] == 0) {
                        queue.offer(neighbor);
                    }
                }
            }
            step++;
        }
        return count == N ? step : -1;
    }
}