题目：

There are a total of numCourses courses you have to take, labeled from 0 to numCourses-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, is it possible for you to finish all courses?

 

Example 1:

Input: numCourses = 2, prerequisites = [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: numCourses = 2, prerequisites = [[1,0],[0,1]]
Output: false
Explanation: There are a total of 2 courses to take. 
             To take course 1 you should have finished course 0, and to take course 0 you should
             also have finished course 1. So it is impossible.

 

Constraints:

• The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
• You may assume that there are no duplicate edges in the input prerequisites.
• 1 <= numCourses <= 10^5

 

代码：

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        if(prerequisites.size() < 1 || numCourses < 1)
            return true;
        
        vector<vector<int>> graph(numCourses);
        vector<int> indegree(numCourses);
        for(int i = 0; i<prerequisites.size(); i++)
        {
            pair<int, int> cur_pair = prerequisites[i];
            if(cur_pair.first == cur_pair.second)
                return false;
            graph[cur_pair.second].push_back(cur_pair.first);
            indegree[cur_pair.first]++;
        }
        
        int count = 0;
        queue<int> q;
        for(int i = 0; i<numCourses; i++)
        {
            if(indegree[i] == 0)
                q.push(i);
        }
        
        while(!q.empty())
        {
            int cur = q.front();
            count++;
            for(auto v : graph[cur])
            {
                indegree[v]--;
                if(indegree[v] == 0)
                    q.push(v);
            }
            q.pop();
        }
        
        return count == numCourses;
    }
};

 

解法效率优于100.00%~！^0^