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LeetCode 210. Course Schedule II

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Question

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, 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: 

1
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] 

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

Hints: This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort. Topological sort could also be done via BFS.

Explanation

这道题是典型的可以用BFS的题，可以将这些关系考虑为一个graph。我们需要一个数据结构来存度数，用一个Queue来先存下没有先修课的课，然后按照度数来一次调出。得到最终的结构。

Code

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public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        HashMap<Integer, Set<Integer>> graph = new HashMap<>();
        int[] indegree = new int[numCourses];
        for (int i=0; i<prerequisites.length; i++) indegree[prerequisites[i][0]]++;
        int[] order = new int[numCourses];
        int index = 0;
        Queue<Integer> queue = new LinkedList<Integer>();
        for (int i=0; i< numCourses; i++) {
            if (indegree[i] == 0) {
                order[index++] = i;
                queue.offer(i);
            }
        }
        while ( !queue.isEmpty()) {
            int prerequisite = queue.poll();
            for (int i=0; i<prerequisites.length; i++) {
                if (prerequisites[i][1] == prerequisite) {
                    indegree[prerequisites[i][0]] --;
                    if(indegree[prerequisites[i][0]] == 0) {
                        order[index++] = prerequisites[i][0];
                        queue.offer(prerequisites[i][0]);
                    }
                }
            }
        }
        return (index == numCourses) ? order : new int[0];
    }
}