210. Course Schedule II (Medium)

There are a total ofncourses you have to take, labeled from0ton - 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:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

Solution 1: BFS, Topological Sort O(V + E); O(V)

    /**
     * BFS O(V + E);O(V)
     * 1.Convert graph presentation from edges to indegree of adjacent list.
     * 2.Add the course with 0 in-degree, which means no prerequisites to the order and queue.
     * 3.While queue is not empty:
     * | The head node dequeue
     * | Then remove it from the graph by reducing the in-degree of its adjacent nodes.
     * | If adjacent node's in-degree becomes 0, add it to the order and queue.
     * Finally, check whether all nodes are visited.
     */
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        if (numCourses <= 0) {
            return null;
        }

        int[] indegree = new int[numCourses];
        int[] order = new int[numCourses];
        int index = 0;
        //1.Convert graph presentation from edges to indegree of adjacent list.
        for (int i = 0; i < prerequisites.length; i++) {
            indegree[prerequisites[i][0]]++;
        }
        //2.Add the course which has no prerequisites to the order.
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0) {
                order[index++] = i;
                queue.offer(i);
            }
        }
        while (!queue.isEmpty()) {
            int pre = queue.poll();
            for (int i = 0; i < prerequisites.length; i++) {
                if (prerequisites[i][1] == pre) {
                    int curCourse = prerequisites[i][0];
                    indegree[curCourse]--;
                    if (indegree[curCourse] == 0) {
                        order[index++] = curCourse;
                        queue.offer(curCourse);
                    }
                }
            }
        }
        return index == numCourses ? order : new int[0];
    }