【LeetCode】210. Course Schedule II

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].

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

分析：
这道题是Course Schedule的延伸，要求我们得到上课的顺序。我们可以继续使用上一道题的算法。
设一个保存上课顺序的向量result。然后在每一次将入度为0的结点入队的同时，将该结点保存在向量result中。在队列为空时，判断向量的长度，若向量长度等于结点的长度，则表示有向图无环，可以完成全部课程，返回向量result；若向量长度不等于结点长度，则表示有向图有环，无法完成全部课程，将向量result清空后再返回。

代码：

class Solution {public:    vector<int> findOrder(int numCourses, vector<pair<int, int> >& prerequisites) {        int* indegree = new int[numCourses];        memset(indegree,0,numCourses*sizeof(int));        queue<int> q;        vector<int> result;        int ss;        for(int i = 0;i<prerequisites.size();i++)        {            ss = prerequisites[i].first;            indegree[ss]++;        }        for(int i = 0;i<numCourses;i++)        {            if(indegree[i] == 0)            {                q.push(i);                result.push_back(i);            }        }        while(!q.empty())        {            int temp = q.front();            q.pop();            for(int i = 0;i<prerequisites.size();i++)            {                if(prerequisites[i].second ==temp)                {                    ss = prerequisites[i].first;                    indegree[ss]--;                    if(indegree[ss] == 0)                    {                        q.push(ss);                        result.push_back(ss);                    }                }            }        }        delete [] indegree;        if(result.size()==numCourses)        return result;        else         {            result.clear();            return result;        }    }};