LeetCode #207 - Course Schedule

最新推荐文章于 2022-11-01 19:09:45 发布

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最新推荐文章于 2022-11-01 19:09:45 发布

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本文介绍了一种算法，用于判断在给定课程及其先修条件的情况下，是否有可能完成所有课程。通过构建图的邻接表并使用拓扑排序的方法，从没有先修要求的课程开始遍历，最终确定所有课程是否都能被完成。

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题目描述：

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?

Example 1:

Input: 2, [[1,0]] 
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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.
You may assume that there are no duplicate edges in the input prerequisites.

这道题其实是需要判断一个图是否为连通图，从入度为零的点出发，也就是从没有先修要求的课程开始，判断是否能够遍历到图中所有的点。因此需要先根据给定的课程先修要求构建图的邻接表，并且记录每个点的入度，然后把入度为零的点加入队列，依次遍历它们的邻接点，同时每次访问一个节点，就把它们的入度减一，如果此时入度为零就可以加入队列。当队列为空时，即在没有入度为零的点，判断此时是否还有点没有访问到即可。

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> course;
        vector<vector<int>> graph(numCourses,course);
        vector<int> indegree(numCourses,0);
        queue<int> start;
        for(int i=0;i<prerequisites.size();i++)
        {
            graph[prerequisites[i].first].push_back(prerequisites[i].second);
            indegree[prerequisites[i].second]++;
        }
        for(int i=0;i<numCourses;i++) if(indegree[i]==0) start.push(i);
        
        while(!start.empty())
        {
            int i=start.front();
            for(int j=0;j<graph[i].size();j++)
            {
                indegree[graph[i][j]]--;
                if(indegree[graph[i][j]]==0) start.push(graph[i][j]);
            }
            start.pop();
        }
        for(int i=0;i<numCourses;i++) if(indegree[i]>0) return false;
        return true;
    }
};