Course Schedule II

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:

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

click to show more hints.


  1. 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.
  2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
  3. Topological sort could also be done via BFS.


时间复杂度 O(n),空间复杂度 O(n)

  1. 先遍历一遍课程,统计所有课的前置课程数量,并把数量为0的加入结果,并入队。index++
  2. 把队里的课程poll出来,讲他们的后制课程的indegree--,此时如果有课程的indegree变为0,则加入结果,入队。index++
  3. 如果index与课程数一样则返回结果,不一样则题目出错,返回new int[0]。




public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        if (numCourses == 0) {
            return null;
        int[] result = new int[numCourses];
        int[] indegree = new int[numCourses];
        int index = 0;
        //calculate how many courses are ahead
        for (int i = 0; i < prerequisites.length; i++) {
        Queue<Integer> queue = new LinkedList<>();
        // Add the course to the order because it has no prerequisites.
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0) {
                result[index++] = i;
        while (!queue.isEmpty()) {
            int pre = queue.poll();
            // indegree of those courses whose pre has been added should reduce;
            for (int i = 0; i < prerequisites.length; i++) {
                if (prerequisites[i][1] == pre) {
                    // if the indegree of current course is 0, it should be added now
                    if (indegree[prerequisites[i][0]] == 0) {
                        result[index++] = prerequisites[i][0];
        return index == numCourses ? result : new int[0];


public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        if(prerequisites == null){
            throw new IllegalArgumentException("illegal prerequisites array");

        int len = prerequisites.length;

        if(numCourses == 0 || len == 0){
            return true;

        //track visited courses
        int[] visit = new int[numCourses];

        // use the map to store what courses depend on a course 
        HashMap<Integer,ArrayList<Integer>> map = new HashMap<Integer,ArrayList<Integer>>();
        for(int[] a: prerequisites){
                ArrayList<Integer> l = new ArrayList<Integer>();
                map.put(a[1], l);

        for(int i=0; i<numCourses; i++){
            if(!canFinishDFS(map, visit, i))
                return false;

        return true;

    private boolean canFinishDFS(HashMap<Integer,ArrayList<Integer>> map, int[] visit, int i){
            return false;
            return true;

            for(int j: map.get(i)){
                if(!canFinishDFS(map, visit, j)) 
                    return false;


        return true;

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