[LeetCode] 210. Course Schedule II 课程清单之二

 

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.

Example 1:

Input: 2, [[1,0]] 
Output: 
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][0,1] .

Example 2:

Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: 
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 . Another correct ordering is [0,1,2,3] or [0,2,1,3][0,1,2,3][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.

Hints:

1. This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
2. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

 

这题是以前那道 Course Schedule 的扩展，那道题只让咱们判断是否能完成全部课程，即检测有向图中是否有环，而这道题咱们得找出要上的课程的顺序，即有向图的拓扑排序 Topological Sort，这样一来，难度就增长了，可是因为咱们有以前那道的基础，而此题正是基于以前解法的基础上稍加修改，咱们从 queue 中每取出一个数组就将其存在结果中，最终如有向图中有环，则结果中元素的个数不等于总课程数，那咱们将结果清空便可。代码以下：

 

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> res;
        vector<vector<int> > graph(numCourses, vector<int>(0));
        vector<int> in(numCourses, 0);
        for (auto &a : prerequisites) {
            graph[a.second].push_back(a.first);
            ++in[a.first];
        }
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] == 0) q.push(i);
        }
        while (!q.empty()) {
            int t = q.front();
            res.push_back(t);
            q.pop();
            for (auto &a : graph[t]) {
                --in[a];
                if (in[a] == 0) q.push(a);
            }
        }
        if (res.size() != numCourses) res.clear();
        return res;
    }
};

 

Github 同步地址：

https://github.com/grandyang/leetcode/issues/210

 

相似题目：

Minimum Height Trees

Course Schedule

Course Schedule III

Alien Dictionary 

Sequence Reconstruction

 

参考资料：

https://leetcode.com/problems/course-schedule-ii/

https://leetcode.com/problems/course-schedule-ii/discuss/59330/Concise-JAVA-solution-based-on-BFS-with-comments

https://leetcode.com/problems/course-schedule-ii/discuss/59342/Java-DFS-double-cache-visiting-each-vertex-once-433ms

 

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