210. Course Schedule II ---- M

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

• For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

 

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: [0,1]
Explanation: 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].

Example 2:

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: 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].

Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]

 

Constraints:

• 1 <= numCourses <= 2000
• 0 <= prerequisites.length <= numCourses * (numCourses - 1)
• prerequisites[i].length == 2
• 0 <= ai, bi < numCourses
• ai != bi
• All the pairs [ai, bi] are distinct.

A:

------------就是  Toplogical sort ( 具体思路，可以看CLIS ）

class Solution {

public:

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

vector<int> res;

vector<int> Color(numCourses, 0);

unordered_map<int, vector<int>> M;

for(auto pre : prerequisites)

M[pre[0]].push_back(pre[1]);

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

if(not dfs(M, Color, res, i))

return {};

}

return res;

}

private:

bool dfs(unordered_map<int, vector<int>>& M, vector<int>& Color, vector<int> & res, int seed){

if(Color[seed] == 2)

return true;

if(Color[seed] == 1)

return false;

// now Color[seed] == 0, meaning unvisited,

Color[seed] = 1;

for(auto v : M[seed]){

if(not dfs(M, Color, res, v)){

return false;

}

}

Color[seed] = 2;

res.push_back(seed); // CLIS says we should insert at front

return true;

}

};

 -------第一遍，  因为不是tree 格式，所以我们用数组做-------------

------------each round, extract the index with 0 out-going dependence, ------------ PASS, but still a little bit slow ----------less memory -------------------

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        vector<int> res;
        vector<int> outCount(numCourses, 0); // number of out dependence
        for(auto& V:prerequisites )
            ++outCount[V[0]];
        
        bool foundLastRound = true;
        while(foundLastRound)
        {
            foundLastRound = false;
            for(int i =0;i<numCourses;++i)
            {
                if(outCount[i] ==0)
                {
                    res.push_back(i);
                    outCount[i] = -1; // will never be called
                    foundLastRound = true;
                    break;
                }
            }
            if(foundLastRound)
            {
                for(auto& V:prerequisites )
                    if(V[1] == res.back())
                        --outCount[V[0]];
            }
        }
        if(res.size()<numCourses)
            res.clear();
        return res;
    }
};

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