Course Schedule II

最新推荐文章于 2025-08-22 18:00:28 发布

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最新推荐文章于 2025-08-22 18:00:28 发布

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分类专栏： Leetcode 文章标签：图论

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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实现。通过构建有向图并利用深度优先搜索来检测环路，确保课程之间的依赖关系得到正确处理。文章提供了一个具体的Java代码示例，展示了如何找出合法的课程完成顺序。

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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][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].

在Course Schedule 1的基础上输出一个可以完成所有课程的序列。只需要在递归完之后记录节点就可以了。代码如下：


public class Solution {
    boolean[] onStack;
    boolean[] isVisited;
    LinkedList<Integer> list;
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] result = new int[numCourses];
        list = new LinkedList<Integer>();
        List<Integer>[] graph = new List[numCourses];
        onStack = new boolean[numCourses];
        isVisited = new boolean[numCourses];
        for(int i = 0; i < prerequisites.length; i++) {
            int key = prerequisites[i][1];
            int value = prerequisites[i][0];
            if(graph[key] == null) graph[key] = new ArrayList<Integer>();
            graph[key].add(value);
        }
        for(int i = 0; i < numCourses; i++) {
            if(isVisited[i] == false && hasCycle(i, graph) == true)
                return new int[0];
        }
        for(int i = 0; i < list.size(); i++) {
            result[i] = list.get(i);
        }
        return result;
    }

    public boolean hasCycle(int i, List<Integer>[] graph) {
        boolean cycle = false;
        onStack[i] = true;
        isVisited[i] = true;
        if(graph[i] != null) {
            for(int c : graph[i]) {
                if(isVisited[c] == false) {
                    cycle = hasCycle(c, graph);
                    if(cycle == true)
                        break;
                } else if(onStack[c] == true) {
                    cycle = true;
                    break;
                }
            }
        }
        onStack[i] = false;
        list.addFirst(i);
        return cycle;
    }
}