Course Schedule

最新推荐文章于 2022-02-16 14:03:41 发布

powerx_yc

最新推荐文章于 2022-02-16 14:03:41 发布

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原文链接：http://www.cnblogs.com/codingEskimo/p/7006970.html

本文介绍了一种利用拓扑排序判断课程先修顺序是否构成有效学习路径的方法。通过构建课程依赖关系的图结构，使用队列实现拓扑排序，判断所有课程是否可以按特定顺序完成。此算法适用于课程规划等依赖关系分析场景。

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Note:

1) For integer node 0~n, we don't need a hashmap, we can just use array to act like a map;

2) How to initial List[] is very important

3) We need two things. One is the relationship between the node and its dependency (out degree map). The other one is dependend nodes' in-degree map.

public class Solution {
    /**
     * @param numCourses a total of n courses
     * @param prerequisites a list of prerequisite pairs
     * @return true if can finish all courses or false
     */
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        // Write your code here
        //Out Degree relationship
        List[] edges = new ArrayList[numCourses]; 
        //In Degree map
        int[] degrees = new int[numCourses];
        
        for (int i = 0; i < numCourses; i++) {
            edges[i] = new ArrayList<Integer>();
        }
        
        for (int i = 0; i < prerequisites.length; i++) {
            degrees[prerequisites[i][0]]++;
            edges[prerequisites[i][1]].add(prerequisites[i][0]);
        }
        
        Queue<Integer> q = new LinkedList<Integer>();
        for (int i = 0; i < numCourses; i++) {
            if (degrees[i] == 0) {
                q.offer(i);
            }
        }
        
        int count = 0;
        while (!q.isEmpty()) {
            int course = q.poll();
            count++;
            int n = edges[course].size();
            for (int i = 0; i < n; i++) {
                int advancedCourse = (int)edges[course].get(i);
                if (degrees[advancedCourse] == 1) {
                    q.offer(advancedCourse);
                }
                degrees[advancedCourse]--;
            }
        }
        return (count == numCourses);
    }
}

转载于:https://www.cnblogs.com/codingEskimo/p/7006970.html