Course Schedule II

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最新推荐文章于 2024-07-14 11:39:13 发布

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分类专栏： breath-first-search topological sort

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本文深入探讨了课程调度算法，通过解析CourseScheduleII问题，阐述了解决多个课程及其先修课程顺序排列的方法。文章详细介绍了如何通过深度优先搜索（DFS）或广度优先搜索（BFS）进行拓扑排序，确保所有课程能够顺利完成。并通过实例展示了不同课程数量和先修关系下的正确课程顺序排列，为解决实际教学计划提供了有效策略。

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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.

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.

The solution of this problem is just a little different from the course schedule 1. If you have problems, please refer to the course schedule 1.

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] ret = new int[numCourses];
        Queue<Integer> queue = new LinkedList<Integer>();
        if (numCourses <= 0)
            return ret;
        
        int j = 0;
        
        int[] counter = new int[numCourses];
        int len = prerequisites.length;
        
        for (int i = 0; i < len; i++) {
            counter[prerequisites[i][0]]++;
        }
        
        for (int i = 0; i < numCourses; i++) {
            if (counter[i] == 0) {
                ret[j++] = i;
                queue.add(i);
            }
        }
        
        while (!queue.isEmpty()) {
            int top = queue.poll();
            for (int i = 0; i < len; i++) {
                if (prerequisites[i][1] == top) {
                    counter[prerequisites[i][0]]--;
                    if (counter[prerequisites[i][0]] == 0) {
                        queue.add(prerequisites[i][0]);
                        ret[j++] = prerequisites[i][0];
                    }
                        
                }
            }
        }
        if (j != numCourses)
            return new int[]{};
        return ret;
    }
}