[LeetCode] 210. Course Schedule II 课程安排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.

Hints:

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

207. Course Schedule的拓展，解法与其类似，这里要按顺序添加要完成的课程。拓扑排序，最后加一步判断是否存在环，如果存在环则返回空集合。使用BFS和DFS均可，区别在于是按照入度还是出度来考虑。

C++：

class Solution {

public:

    /**

     * 完成所有的课程的顺序

     * bfs拓扑排序

     * @param numCourses 课程数量

     * @param prerequisites 课程先序关系

     * @return 能完成返回课程顺序,否则返回空

     */

    vector<int> findOrder(int numCourses, vector<pair<int, int> >& prerequisites) {

        vector<int> heads(numCourses, -1), degree(numCourses, 0), points, args;

        pair<int, int> p;

        int from, to, count = 0, len = prerequisites.size();

        /* 构造有向图,邻接表 */

        for (int i = 0; i < len; ++i) {

            p = prerequisites[i];

            from = p.second;

            to = p.first;

            ++degree[to];

            args.push_back(heads[from]);

            points.push_back(to);

            heads[from] = count++;

        }   

        /* bfs拓扑排序,依次移除入度为0的点 */

        vector<int> ret;

        queue<int> q;

        for (int i = 0; i < numCourses; ++i)

            if (degree[i] == 0) q.push(i);

        while (!q.empty()) {

            from = q.front();

            ret.push_back(from); // 课程完成，添加到结果集中

            q.pop();

            to = heads[from];

            while (to != -1) {

                if(--degree[points[to]] == 0) q.push(points[to]);

                to = args[to];

            }

        }   

        /* 判定是否所有的点入度都为0,若是则不存在环,否则存在环 */

        for (int i = 0; i < numCourses; ++i)

            if (degree[i] > 0) {

                ret.clear();

                break;

            }   

        return ret;

    }

};

C++：DFS

class Solution {

public:

    /**

     * 完成所有的课程的顺序

     * dfs拓扑排序

     * @param numCourses 课程数量

     * @param prerequisites 课程先序关系

     * @return 能完成返回课程顺序,否则返回空

     */

    vector<int> findOrder(int numCourses, vector<pair<int, int> >& prerequisites) {

        vector<int> heads(numCourses, -1), degree(numCourses, 0), points, args;

        pair<int, int> p;

        int from, to, count = 0, len = prerequisites.size();

        /* 构造有向图,邻接表 */

        for (int i = 0; i < len; ++i) {

            p = prerequisites[i];

            from = p.second;

            to = p.first;

            ++degree[from];

            args.push_back(heads[to]);

            points.push_back(from);

            heads[to] = count++;

        }   

        /* dfs拓扑排序,依次移除出度为0的点 */

        vector<int> ret;

        queue<int> q;

        for (int i = 0; i < numCourses; ++i)

            if (degree[i] == 0) q.push(i);

        while (!q.empty()) {

            to = q.front();

            ret.push_back(to); // 课程完成添加到结果集中

            q.pop();

            from = heads[to];

            while (from != -1) {

                if(--degree[points[from]] == 0) q.push(points[from]);

                from = args[from];

            }

        }   

        /* 判定是否所有的点入度都为0,若是则不存在环,否则存在环 */

        for (int i = 0; i < numCourses; ++i)

            if (degree[i] > 0) {

                ret.clear();

                break;

        }   

        /* 逆序 */

        reverse(ret.begin(), ret.end());

        return ret;

    }

};

巴特西

[LeetCode] 210. Course Schedule II 课程安排II

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