课程表 II

Problem

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

• For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]

Output: [0,1]

Explanation: 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].

Example 2:

Input: numCourses = 4, prerequisites = [[1,0], [2,0], [3,1], [3,2]]

Output: [0,2,1,3]

Explanation: 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].

Example 3:

Input: numCourses = 1, prerequisites = []

Output: [0]

Constraints:

• 1 <= numCourses <= 2000
• 0 <= prerequisites.length <= numCourses * (numCourses - 1)
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• ai != bi
• All the pairs [ai, bi] are distinct.

Idea

Model the courses as a directed graph with an edge bi → ai for each prerequisite, and find a topological order with Kahn's algorithm (BFS).

Count the in-degree of every course and push all courses with in-degree 0 into a queue. Repeatedly pop a course, append it to the order, and decrease the in-degree of its neighbors; any neighbor whose in-degree drops to 0 is pushed into the queue.

If the order contains all numCourses courses, return it. Otherwise the graph has a cycle, so return an empty array.

Time complexity: \(O(V + E)\), Space complexity: \(O(V + E)\), where \(V\) is numCourses and \(E\) is the number of prerequisites

Pseudo Code

graph = array of numCourses empty lists;
indegree = array of numCourses zeros;
for ([a, b] in prerequisites) {
    graph[b].add(a);
    indegree[a]++;
}

queue = empty queue;
for (i = 0; i < numCourses; i++) {
    if (indegree[i] == 0) {
        queue.push(i);
    }
}

order = empty list;
while (!queue.isEmpty()) {
    course = queue.pop();
    order.add(course);
    for (next in graph[course]) {
        indegree[next]--;
        if (indegree[next] == 0) {
            queue.push(next);
        }
    }
}

if (order.size() == numCourses) {
    return order;
}
return [];