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.
拓扑排序。用一个队列存入度为0的节点,依次出队,将与出队节点相连的节点的入度减1,如果入度减为0,将其放入队列中,直到队列为空。如里最后还有入度不为0的节点的话,说明有环,否则无环。
1 class Solution { 2 public: 3 bool canFinish(int numCourses, vector<vector<int>>& prerequisites) { 4 vector<vector<int>> graph(numCourses, vector<int>(0)); 5 vector<int> inDegree(numCourses, 0); 6 for (auto u : prerequisites) { 7 graph[u[1]].push_back(u[0]); 8 ++inDegree[u[0]]; 9 } 10 queue<int> que; 11 for (int i = 0; i < numCourses; ++i) { 12 if (inDegree[i] == 0) que.push(i); 13 } 14 while (!que.empty()) { 15 int u = que.front(); 16 que.pop(); 17 for (auto v : graph[u]) { 18 --inDegree[v]; 19 if (inDegree[v] == 0) que.push(v); 20 } 21 } 22 for (int i = 0; i < numCourses; ++i) { 23 if (inDegree[i] != 0) return false; 24 } 25 return true; 26 } 27 };