[leetcode]207. Course Schedule

 第一种图的方法：超时

struct Edge{
    int to;
    Edge(int x){
        to = x;
    }
};
struct Node{
    vector<Edge> edges;
    int indegree;
    Node(){
        indegree=0;
    }
};
struct Graph{
    Node * node;
    int num;
    Graph(int n){
        node =new Node[n];
        num = n;
    }
    void insert(const pair<int,int> & p){
        node[p.first].edges.push_back(Edge(p.second));
        node[p.second].indegree++;
    }
    ~Graph(){
        delete [] node;
    }
};
class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        Graph graph(numCourses);
        for(auto item : prerequisites){
            graph.insert(item);
        }
        bool visited[numCourses];
        queue<int> q;
        for(int i = 0;i < numCourses; i++){
            visited[i]= false;
            if(graph.node[i].indegree == 0){
                q.push(i);
                visited[i]= true;
            }
        }
        while(!q.empty()){
            int index = q.front();
            q.pop();
            for(auto item : graph.node[index].edges){
                graph.node[item.to].indegree--;
            }
            for(int i = 0;i < numCourses; i++){
                if(graph.node[i].indegree == 0){
                    visited[i] = true;
                    q.push(i);
                }
            }
        }
        for(int i = 0;i < numCourses; i++){
            if(!visited[i]){
                return false;
            }
        }
        return true;
        
    }
};

第二种方法：

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        int indegree[numCourses];
        bool visited[numCourses];
        queue<int> q;
        unordered_map<int,vector<int>> m;
        for(int i=0;i<numCourses;i++){
            indegree[i]=0;
            visited[i] = false;
        }
        for(auto item : prerequisites){
            m[item.first].push_back(item.second);
            indegree[item.second]++;
        }
        for(int i=0;i<numCourses;i++){
            if(indegree[i] == 0){
                q.push(i);
                visited[i] = true;     
            }
        }
        while(!q.empty()){
            int index = q.front();
            q.pop();
            // auto p = m.find(index);
            for(auto item : m[index]){
                indegree[item]--;
                if(indegree[item] == 0){
                    q.push(item);
                    visited[item] = true;
                }
            }
        }   
        for(int i=0;i<numCourses;i++){
           if(!visited[i]) 
               return false;
        }
        return true;
    }
};

 思路是拓扑排序，复杂度O(n+e)