LeetCode C++ 210. Course Schedule II（思路）

There are a total of numCourses courses you have to take, labeled from 0 to numCourses - 1. You are given an array prerequisites where prerequisites[i] = [ai, bi] indicates that you must take course bi first if you want to take course ai.

For example, the pair [0, 1], indicates that to take course 0 you have to first take course 1.

Return the ordering of courses you should take to finish all courses. If there are many valid answers, return any of them. If it is impossible to finish all courses, return an empty array.

Example 1:

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

Input: numCourses = 4, prerequisites = [[1,0],[2,0],[3,1],[3,2]]
Output: [0,2,1,3]
Explanation: There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0.
So one correct course order is [0,1,2,3]. Another correct ordering is [0,2,1,3].
Example 3:

Input: numCourses = 1, prerequisites = []
Output: [0]
 

Constraints:

1 <= numCourses <= 2000
0 <= prerequisites.length <= numCourses * (numCourses - 1)
prerequisites[i].length == 2
0 <= ai, bi < numCourses
ai != bi
All the pairs [ai, bi] are distinct. 

解法1  [DFS] 拓扑排序

class Solution {
private:
    bool isCyclic(const vector<vector<int>>& G, int u, vector<bool>& visited, vector<bool>& recStack, stack<int>& topStack) {
         //已经访问过
         //当前顶点加入递归栈中
         //没有访问过
          //存在环
           //访问过v且v在此时的递归栈中,存在回边
          //当前顶点移出递归栈
          //访问完当前顶点的所有邻接点后入栈当前顶点
          //没有环
    } 
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
          //存在环
           //无法完成课程
         //不断弹出栈中的顶点
             //得到拓扑序列
           //不存在环,可以完成课程
    
}; 

解法2 BFS拓扑排序

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
          //拓扑序列
         
          //入度+1
        
    }
};