leetcode 210. Course Schedule II

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].

与Course Schedule类似的题目, 使用拓扑遍历！

找到indegree为0的点

class Solution {
public:
    int findpoint(vector<int> &indegree,int num)
    {
        for(int i=0;i<num;i++)
        {
            if(indegree[i]==0)
            {
                indegree[i]--;
                return i;
            }
        }
        return -1;
    }
    
    vector<int> findOrder(int num, vector<pair<int, int>>& prer) 
    {
        sort(prer.begin(), prer.end());     //这三行代码是去重三剑客
        auto it = unique(prer.begin(), prer.end());
        prer.resize(it - prer.begin());
        
        //形成邻接矩阵
        vector<int> indegree(num, 0);
        vector<vector<int>> p(num, indegree);
       
        for (auto it : prer)
        {
            p[it.first][it.second] = 1;
            indegree[it.second]++;
        }
       
        //就是看一个有向图是否形成环，return 0。
        int taken = 0;
        vector<int> ret;
        while (taken != num)
        {
            int k = findpoint(indegree, num);
            if(k == -1)    //没有入度为0的点
            {
                ret.clear();
                return ret;
            }
            else
                ret.push_back(k);
            
            for (int i = 0; i < num; i++)        
            {  
                if( p[k][i] == 1)
                    indegree[i]--;
            }
            taken++;    
        }
        reverse(ret.begin(),ret.end()); //输出结果需要反转
        return ret;
    }
};