207. Course Schedule

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.

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给出数字n代表有n门课，编号0~n-1，一些课是有要求要先上一些别的课，问是否能修完全部课。这里能修完就相当于需求图中没有回环，所以可以用拓扑排序的方法解决问题。先找一个入度为0的节点，把它摘除，然后他指向的节点的入度就减一，减一的过程中如果有节点的入度变为0，就入栈，方便下次的摘除；然后循环摘除节点，直到没有入度为0的节点为止。循环过程中进行计数，如果摘除的节点个数小于所有节点数，说明图中有环，返回false，否则返回true。

代码：

class Solution{public:bool canFinish(int numCourses, vector<pair<int, int> >& prerequisites) {vector<vector<int> > graph(numCourses);vector<int> in_cnt(numCourses);for(int i = 0; i < prerequisites.size(); ++i){graph[prerequisites[i].first].push_back(prerequisites[i].second);in_cnt[prerequisites[i].second]++;}stack<int> stk;int cnt = 0;for(int i = 0; i < numCourses; ++i){if(in_cnt[i] == 0) stk.push(i);}while(!stk.empty()){int cur = stk.top();stk.pop();++cnt;for(int i = 0; i < graph[cur].size(); ++i){if(--in_cnt[graph[cur][i]] == 0){stk.push(graph[cur][i]);}}}return cnt >= numCourses;}};