210. Course Schedule II

问题：

给定编号为0～numCourses-1 的课程，以及课程的依赖关系prerequisites

其中prerequisites[i] = [ai, bi]，ai之前必须先上了bi的课程。 bi->ai

若能够上完所有的课程，给出一个上课顺序。

否则，返回空数组。

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：BFS

有向图遍历，判断是否有内循环♻️ 环。有，则不符合要求。

根据依赖关系prerequisites，构造有向图：graph[node1] = {node2, node3...noden} 

  • node1 -> node2
  • node1 -> node3
  • ...
  • node1 -> noden

同时记录有向信息，入度数：indegree[node1]=x(count{node2,node3...noden})

 

  • 从入度为0的节点开始遍历图。
    • queue.push(nodex(where indegree[nodex]==0))
  • for
    • 一个个pop节点：cur（同时该节点被遍历，加入res）
    • 对cur的所有后续节点graph[cur]: nextn，indegree[nextn]--
      • 若后续节点nextn的入度==0，加入queue：queue.push(nextn)
    • 继续pop下一个节点。
  • 最后res.size!=课程数numCourses，则说明没有遍历完有向图，
    • 返回空数组。
    • 其他，返回res。

 

代码参考：

 1 class Solution {
 2 public:
 3     vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
 4         vector<vector<int>> graph(numCourses);
 5         vector<int> res;
 6         queue<int> q;
 7         vector<int> indegree(numCourses, 0);
 8         for(auto line:prerequisites) {
 9             graph[line[1]].push_back(line[0]);
10             indegree[line[0]]++;
11         }
12         for(int i=0; i<numCourses; i++) {
13             //indegree==0 node has only output.
14             if(indegree[i]==0) q.push(i);
15         }
16         while(!q.empty()) {
17             int sz = q.size();
18             for(int i=0; i<sz; i++) {
19                 int cur = q.front();
20                 q.pop();
21                 res.push_back(cur);
22                 for(auto nextn:graph[cur]) {
23                     indegree[nextn]--;
24                     if(indegree[nextn]==0) q.push(nextn);
25                 }
26             }
27         }
28         if(res.size()!=numCourses) return vector<int>();
29         else return res;
30     }
31 };

 

posted @ 2021-02-28 10:08  habibah_chang  阅读(34)  评论(0)  编辑  收藏  举报