210. Course Schedule II

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最新推荐文章于 2025-08-20 09:37:41 发布

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分类专栏： leetcode 文章标签： leetcode

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本文介绍了一种用于解决课程先修顺序问题的算法。该算法通过构建图结构并利用拓扑排序来确定合理的课程学习顺序。文章提供了完整的代码实现，并通过两个示例展示了如何使用此算法来确定有效的课程学习路径。

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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].

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[][] matrix = new int[numCourses][numCourses];
        int[] indegree = new int[numCourses];
        int[] res = new int[numCourses];
        int index = 0;
        int count = 0;
        for (int i = 0; i < prerequisites.length; i++) {
            int ready = prerequisites[i][0];
            int pre = prerequisites[i][1];
            if (matrix[pre][ready] == 0)
                indegree[ready]++;
            matrix[pre][ready] = 1;
        }
        LinkedList<Integer> queue = new LinkedList<Integer>();
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0)
                queue.offer(i);
        }
        while (!queue.isEmpty()) {
            int course = queue.poll();
            count++;
            res[index++] = course;
            for (int i = 0; i < numCourses; i++) {
                if (matrix[course][i] != 0) {
                    if (--indegree[i] == 0) 
                        queue.offer(i);
                }
            }
        }
        if (count == numCourses) return res;
        return new int[0];
    }
}

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<vector <int> > matrix(numCourses, vector<int>(numCourses, 0));
        vector<int> indegree(numCourses, 0);
        vector<int> res;
        int count = 0;
        for (int i = 0; i < prerequisites.size(); i++) {
            int ready = prerequisites[i].first;
            int pre = prerequisites[i].second;
            if (matrix[pre][ready] == 0)
                indegree[ready]++;
            matrix[pre][ready] = 1;
        }

        queue<int> coursequeue;
        for (int i = 0; i < numCourses; i++) {
            if (indegree[i] == 0)
                coursequeue.push(i);
        }
        while (!coursequeue.empty()) {
            int course = coursequeue.front();
            coursequeue.pop();
            res.push_back(course);
            count++;
            for (int i = 0; i < numCourses; i++) {
                if (matrix[course][i] != 0) {
                    if (--indegree[i] == 0)
                        coursequeue.push(i);
                }
            }
        }
        if (count == numCourses) return res;
        return vector<int>();
        //return res;
    }

};