210. Course Schedule II 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.

Example 1:

Input: 2, [[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: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [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] .

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.

在LeetCode Course Schedule的基础上，要求输出一种拓扑排序结果。
因为已经知道拓扑排序的Kahn解法，即BFS解法，那么输出一种拓扑排序的结果是很简单的事。在BFS的过程中，存储在队列中的点肯定都是入度为0的点，也就是这些课程可以在这一阶段完成，所以我们只需要在每轮BFS的时候，保存一下这一轮的节点就好了。
完整代码如下：

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int> >& prerequisites)
    {
        vector<unordered_set<int> > pre(numCourses), next(numCourses);
        for (const auto& p : prerequisites) {
            pre[p.first].insert(p.second);
            next[p.second].insert(p.first);
        }
        vector<int> ans;
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (pre[i].empty())
                q.push(i);
        }
        int edges = prerequisites.size();
        while (!q.empty()) {
            int n = q.size();
            for (int i = 0; i < n; ++i) {
                int cur = q.front();
                ans.push_back(cur);
                q.pop();
                for (const auto& nxt : next[cur]) {
                    pre[nxt].erase(cur);
                    –edges;
                    if (pre[nxt].empty())
                        q.push(nxt);
                }
            }
        }
        if (edges > 0)
            return {};
        else
            return ans;
    }
};

本代码提交AC，用时23MS。

二刷。在上一题的基础上，要求输出拓扑排序的顺序，比之前简单的解法代码，不需要unordered_set：

class Solution {
public:
	vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
		vector<vector<int>> graph(numCourses, vector<int>(numCourses, 0));
		vector<int> indegree(numCourses, 0);
		for (int i = 0; i < prerequisites.size(); ++i) {
			int a = prerequisites[i][0], b = prerequisites[i][1];
			graph[b][a] = 1;
			++indegree[a];
		}
		vector<int> ans;
		while (true) {
			int validid = -1;
			for (int i = 0; i < numCourses; ++i) {
				if (indegree[i] == 0) {
					validid = i;
					--indegree[i];
					break;
				}
			}
			if (validid == -1)break;
			ans.push_back(validid);
			for (int i = 0; i < numCourses; ++i) {
				if (graph[validid][i] == 1) {
					graph[validid][i] = 0;
					--indegree[i];
				}
			}
		}
		if (ans.size() != numCourses)return { };
		return ans;
	}
};

本代码提交AC，用时80MS。