207. Course Schedule

#week17

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, is it possible for you to finish all courses?

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 it is possible.

2, [[1,0],[0,1]]

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.

Note:

1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
2. You may assume that there are no duplicate edges in the input prerequisites.

分析

这一题，很明显，可以看是否能够得到一个拓扑序列，就可以看能够完成

其实也就是检验是否是一个有环的图，如果有环就没法完成课程

如果无环则可以

所以检验有环无环，就可以通过求拓扑或者DFS等方式完成

题解

这个是摘要别人的BFS代码，我这个写的有点多有点丑：

class Solution {
public:
    bool canFinish(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<unordered_set<int>> graph = make_graph(numCourses, prerequisites);
        vector<int> degrees = compute_indegree(graph);
        for (int i = 0; i < numCourses; i++) {
            int j = 0;
            for (; j < numCourses; j++)
                if (!degrees[j]) break;
            if (j == numCourses) return false;
            degrees[j] = -1;
            for (int neigh : graph[j])
                degrees[neigh]--;
        }
        return true;
    }
private:
    vector<unordered_set<int>> make_graph(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<unordered_set<int>> graph(numCourses);
        for (auto pre : prerequisites)
            graph[pre.second].insert(pre.first);
        return graph;
    }
    vector<int> compute_indegree(vector<unordered_set<int>>& graph) {
        vector<int> degrees(graph.size(), 0);
        for (auto neighbors : graph)
            for (int neigh : neighbors)
                degrees[neigh]++;
        return degrees;
    }
};