【leetcode】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].

解题思路：

根据依赖关系建立映射表(用一个map存储，key为课程，value是一个集合，表示当前课程所依赖的课程们)

然后逐个遍历课程，若存在被依赖，则解除依赖（解除依赖后发现没有依赖关系了，则删除映射表中对应项）

class Solution {

public:

    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {

        vector<int> res;

        if(numCourses<=0) return res;

        if(numCourses==1 || prerequisites.empty()){

            res.push_back(numCourses);

            return res;

        }

        map<int,set<int> > cmap;

        for(int i=0;i<prerequisites.size();i++){

            if(cmap.find(prerequisites[i].first) != cmap.end()){

                cmap[prerequisites[i].first].insert(prerequisites[i].second);

            }

            else{

                set<int> tmp;

                tmp.insert(prerequisites[i].second);

                cmap.insert(pair<int,set<int>>(prerequisites[i].first,tmp));

            }

        }

        for(int i=0;i<numCourses;i++){

            if(cmap.find(i) == cmap.end()){

                res.push_back(i);//找到了一个没有依赖的课程

                for(int j=0;j<cmap.size();j++){//解除所有依赖于此课程的依赖关系

                    set<int> ite=cmap[j];

                    if(ite.find(i) !=ite.end()){//找到了依赖关系

                        ite.erase(i);

                        if(ite.empty()){//解除依赖后，依赖课程没有依赖了，则将其从依赖表中删除

                            cmap.erase(j);

                        }

                    }

                    //当前课程不依赖于本次课程

                }

            }

        }

        //没找到入度为0的节点，说明图中有环，即不可能完成所有的课程

        if(!cmap.empty()){

            res.clear();

        }

        return res;

    }

};

然后并没有AC_(:з」∠)_有一个始终过不去，如下：

Submission Result: Wrong Answer 

Input:1, [ ]Output:[1]Expected:Special judge: No expected output available._(:з」∠)_求各位大神指导