Leetcode-210: Course Schedule II

最新推荐文章于 2025-08-05 16:42:55 发布

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最新推荐文章于 2025-08-05 16:42:55 发布

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分类专栏： leetcode 文章标签： leetcode 深度优先搜索

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本文介绍了一种课程调度算法，用于确定完成所有课程的正确顺序。通过使用深度优先搜索（DFS）的逆序方法来解决该问题，并提供了一个具体的实现案例。

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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].

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.

click to show more hints.

这道题是Course Schedule的后续，比起之前只需要判断是否存在环路，现在还需要返回正确的上课顺序，实际上就是返回DFS的逆顺序，只需要维护一个链表或者栈，在访问了某个节点之后将其加入即可。

class Solution {
    private int GRAY = 1, BLACK = 2, WHITE = 0;
    
    private List<List<Integer>> edges;
    private int[] visited;
    private List<Integer> list = new ArrayList<>();
    
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        if (numCourses <= 0)
            return new int[0];
        
        edges = new ArrayList<>();
        visited = new int[numCourses];
        for (int i = 0; i < numCourses; ++i)
            edges.add(new ArrayList<Integer>());
        
        for (int[] edge : prerequisites) {
            edges.get(edge[0]).add(edge[1]);
        }
        
        for (int from = 0; from < numCourses; ++from) {
            if (!visit(edges, from))
                return new int[0];
        }
        
        int[] res = new int[numCourses];
        for (int i = 0; i < numCourses; ++i) {
            res[i] = list.get(i);
        }
        return res;
    }
    
    private boolean visit(List<List<Integer>> edges, int from) {
        if (visited[from] == GRAY)
            return false;
        if (visited[from] == BLACK)
            return true;
        
        visited[from] = GRAY;
        
        if (!edges.get(from).isEmpty())
            for (Integer to : edges.get(from)){
                if (!visit(edges, to))
                    return false;
            }
        
        visited[from] = BLACK;
        list.add(from);
        return true;
    }
}