[LeetCode 210] Course Schedule II

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: The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

click to show more hints.

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.
• 1. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
• 1. Topological sort could also be done via BFS.

Diffculty
Medium

Similar Problems
[LeetCode ] Course Schedule Medium [LeetCode ] Alien Dictionary Hard [LeetCode ] Minimum Height Tree Medium [LeetCode ] Sequence Reconstruction Medium

Analysis

这道题跟 [Course Schedule][] 相比只改了一句话，把原来的 is it possible for you to finish all courses? 改成了 return the ordering of courses you should take to finish all courses。 原来那道题只让我们判断是否能完成所有课程，即检测有向图中是否有环，而这道题我们得找出要上的课程的顺序，即有向图的拓扑排序，难度略有增加。

由于我们有之前那道的基础，我们可以在之前解法的基础上稍加修改来实现目的，我们从 queue 中每取出一个数组就将其存在结果中，最终若有向图中有环，则结果中元素的个数不等于总课程数，那我们将结果清空即可。 还需注意，prerequisites 为空的时候，任意输出一组结果即可。

Solutions

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);  // 增加这一行
            q.pop();
            for (auto &a : graph[t]) {
                --in[a];
                if (in[a] == 0) q.push(a);
            }
        }
        if (res.size() != numCourses)   // 上面的循环结束后，如果 res.size 跟课程总数不相等，说明有环。
            res.clear();
        return res;
    }
};