Course Schedule II

最新推荐文章于 2021-09-30 21:12:33 发布

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本文介绍了一种解决课程排序问题的算法，通过拓扑排序的方式找出完成所有课程的有效顺序。该算法适用于存在先修课程约束的情况，能够处理多个正确排序的可能性。

题目

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].

分析

该问题是一个一个典型的topological排序问题，实现较为简单，每次将入度为0的节点加入解向量的最后，然后删去该节点，并删去与该节点关联的所有边。直到最后若图已为空，则存在一个拓扑排序，排序为解向量中的顺序，若图不为空，则表示不存在拓扑排序。

在本问题中，用一个indegree向量来存放每个节点的入度，若入度为-1则表示该节点已被删去

代码

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int> >& prerequisites) {
        vector<int> indegree(numCourses,0);
        for (int i = 0;i < prerequisites.size();i++)
        	indegree[prerequisites[i].first]++;
        vector<int> result;
        while (true)
        {
        	bool Continue = false;
        	for (int i = 0;i < indegree.size();i++)
        		if (indegree[i] == 0)
        		{
        			Continue = true;
        			indegree[i] = -1;
        			result.push_back(i);
        			for (int j = 0;j < prerequisites.size();j++)
        				if (prerequisites[j].second == i)
        					indegree[prerequisites[j].first]--;
				}
			if (!Continue)
			{
				for (int i = 0;i < indegree.size();i++)
					if (indegree[i] != -1)
						return vector<int>();
				return result;
			}	  
		}
    }
};