210. Course Schedule II

最新推荐文章于 2022-05-19 09:39:58 发布

BigFatSheep

最新推荐文章于 2022-05-19 09:39:58 发布

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本文介绍了一种解决课程依赖问题的方法，通过使用入度数组存储每门课程的依赖数量，然后找到所有入度为0的课程，将其加入结果数组和队列中。在队列不为空的情况下，不断移除已完成的课程并更新剩余课程的依赖数，直至完成所有课程的学习顺序安排。

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Description

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.

Example 1:

Input: 2, [[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: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [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] .
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.

Problem URL

Solution

给一个数组表示课程之间的依赖关系和一共有多闪门课，问能否上的了这些课。

Using indegree array to store the courses have indegree. Then get the courses which have 0 indegreee and add them into result array and queue.

While queue is not empty, poll out the course, then find it in prerequisites, minus one for courses who have indegree from it. If its indegree becomes zero, add it to result and queue.

If index of result is not equal to the num of courses, return empty array.

Code

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        if (numCourses == 0){
            return new int[0];
        }
        int[] indegree = new int[numCourses];
        int[] result = new int [numCourses];
        int index = 0;
        //find the courses which need prerequisites.
        for (int i = 0; i < prerequisites.length; i++){
            indegree[prerequisites[i][0]]++;
        }
        
        //add the courses which do not need prerequisites.
        Queue<Integer> queue = new LinkedList<>();
        for (int i = 0; i < indegree.length; i++){
            if (indegree[i] == 0){
                queue.offer(i);
                result[index++] = i;
            }
        }
        //add courses to result in the order of queue.
        while(!queue.isEmpty()){
            int course = queue.poll();
            //course polled means a prerequisites course has finished, we could eliminate its indegree
            for (int i = 0; i < prerequisites.length; i++){
                if (prerequisites[i][1] == course){
                    indegree[prerequisites[i][0]]--;
                    //It means we could take this class, add it to queue and result.
                    if (indegree[prerequisites[i][0]] == 0){
                        queue.offer(prerequisites[i][0]);
                        result[index++] = prerequisites[i][0];
                    }
                }
                
            }
        }
        
        return index == numCourses ? result : new int[0];
    }
}