210. Course Schedule II 难度：medium 类别：图_210. course schedule ii (medium)-优快云博客

本文介绍了一种基于拓扑排序的课程排序算法，旨在解决给定课程及其先修条件下的合理学习顺序问题。通过构建图结构并利用拓扑排序原理，算法能够有效找出所有课程的学习顺序，适用于课程安排等场景。

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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].

思路：

这道题同样是考察拓扑排序，在判断是否能进行拓扑排序的同时输出排序结果，就先按拓扑排序的步骤对图进行排序，最后判断结果序列的元素个数是否等于节点数，是则返回该序列，否则返回空序列。

程序：

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        <img src="https://img-blog.youkuaiyun.com/20160929140739208?watermark/2/text/aHR0cDovL2Jsb2cuY3Nkbi5uZXQv/font/5a6L5L2T/fontsize/400/fill/I0JBQkFCMA==/dissolve/70/gravity/Center" alt="" />vector<int> res;
	vector<vector<int> > adj(numCourses);
	vector<int> indegree(numCourses,0);
	queue<int> q;
	
	for(int i = 0;i < prerequisites.size();i++)
	{
		adj[prerequisites[i].second].push_back(prerequisites[i].first);
		indegree[prerequisites[i].first]++;
	}
	
	for(int i = 0;i < indegree.size();i++)
	{
		if(indegree[i] == 0)
			q.push(i);
	}
	
	while(!q.empty())
	{
		int u = q.front();
		q.pop();
		res.push_back(u);
		
		for(int i = 0;i < adj[u].size();i++)
		{
			indegree[adj[u][i]]--;
			if(indegree[adj[u][i]] == 0)
				q.push(adj[u][i]);
		}
	}
	
	if(res.size() < numCourses)
	{
		res.clear();
		return res;
	}
	
	else
		return res;
        
    }
};