[Leetcode] 210. Course Schedule II 解题报告

题目：

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

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].

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.

思路：

思路完全等同于Leetcode 207，唯一的区别在于需要记录最终的拓扑排序结果。请参考我在《[Leetcode] 207. Course Schedule 解题报告》中的解释和下面的代码片段。

代码：

class Solution {public:    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {        vector<unordered_set<int>> graph(numCourses, unordered_set<int>());          vector<int> indegree(numCourses, 0);          for(auto pre : prerequisites) {                     // initialize the graph              if(graph[pre.second].count(pre.first) == 0) {                  graph[pre.second].insert(pre.first);                  indegree[pre.first]++;              }          }         vector<int> orders;        for(int i = 0, j = 0; i < numCourses; i++) {              for(j = 0; j < numCourses; j++) {               // find the course that can be taken now                  if(indegree[j] == 0) {                      break;                  }              }              if(j == numCourses) {                   return {};              }              orders.push_back(j);            --indegree[j];                                  // update the indegree related to j              for(auto p : graph[j]) {                  --indegree[p];              }          }          return orders;      }};