210. Course Schedule II 图的dfs算法

There are a total of n courses you have to take, labeled from 0 ton - 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. Read more abouthow a graph is represented.

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分析。解法与前一题相同，只需要将每次查找完的点存起来。注意亮点是，被查找过的点不进入存储。以及由于是dfs，最后需要反转。

class Solution {
public:
 bool dfs(int t,vector<int>& visit,vector<vector<int>>& graph,bool& index,vector<int>& result){

        if(visit[t]==1) {return false;}
        if(visit[t]==2) {return true;}
        visit[t]=1;
        for(int i=0;i<graph[t].size();++i)
        {
                if(!dfs(graph[t][i],visit,graph,index,result))
                return false;
        }
        result.push_back(t);
          visit[t]=2;
          return true;
    }
    vector<int> findOrder(int num, vector<pair<int, int>>& pre) {
        vector<int> result;
        vector<int> visit(num,0);
        vector<vector<int>> graph(num);
        for(int i=0;i<pre.size();++i)
        {
            graph[pre[i].second].push_back(pre[i].first);
        }
        bool index=false;
        for(int i=0;i<num;++i)
        {
           
            if(!dfs(i,visit,graph,index,result))
            return vector<int>();
            
        }
        reverse(result.begin(),result.end());
        return result;
    }
};