leetcode Course Schedule

最新推荐文章于 2025-09-13 14:08:34 发布

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最新推荐文章于 2025-09-13 14:08:34 发布

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文章标签： leetcode 算法

本文链接：https://blog.youkuaiyun.com/menghan1224/article/details/52279057

本文探讨了如何判断在存在先修课程约束的情况下，是否能够完成所有课程的问题。通过使用拓扑排序的方法，每次找到并移除入度为0的顶点及其指向的边，直至无法再移除或所有课程都被遍历。

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, is it possible for you to finish all courses?

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 it is possible.

2, [[1,0],[0,1]]

There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

click to show more hints.

拓扑排序问题，每次找到一条入度为0的点，删去该点指向其他节点的边，直到所有点遍历完毕，代码：

public boolean canFinish(int numCourses, int[][] prerequisites) {
    Set[] outset=new Set[numCourses];
    for(int i=0;i<outset.length;i++)
        outset[i]=new HashSet<>();
    for(int[] edge:prerequisites){
        outset[edge[1]].add(edge[0]);
    }
    int[] insize=new int[numCourses];
    for(int i=0;i<numCourses;i++){
        Set temp=outset[i];
        Iterator<Integer> iter=temp.iterator();
        while(iter.hasNext()){
            insize[iter.next()]++;
        }
    }
    for(int i=0;i<numCourses;i++){
        int j=0;
        for(;j<numCourses;j++){
            if(insize[j]==0) break;
        }
        if(j==numCourses) return false;
        insize[j]=-1;
        Set st=outset[j];
        Iterator<Integer> it=st.iterator();
        while(it.hasNext()){
            insize[it.next()]--;
        }
    }
    return  true;
}