210. Course Schedule II

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

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.

You may assume that there are no duplicate edges in the input prerequisites.

Thoughts

给一系列pair代表上第一个元素代表的课前必须先上第二个元素，找出一个合理的上课顺序能上完所有课程，如果不存在则返回空。有先后顺序的图，拓扑排序。依次BFS遍历入度为0的点并把相应的边删去，随着遍历依次把入度为0输出。

Code

/*
 * @lc app=leetcode id=210 lang=cpp
 *
 * [210] Course Schedule II
 */

// @lc code=start
class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        vector<int> in(numCourses, 0);
        vector<unordered_set<int>> edges(numCourses, unordered_set<int>());
        for (const auto &p : prerequisites) {
            edges[p[1]].insert(p[0]);
            ++in[p[0]];
        } 
        queue<int> q;
        for (int i = 0; i < numCourses; ++i) {
            if (in[i] == 0) q.push(i);
        }
        vector<int> res;
        while (!q.empty()) {
            auto t = q.front(); q.pop();
            res.push_back(t);
            for (auto e : edges[t]) {
                --in[e];
                if (in[e] == 0) q.push(e);
            }
        }
        return res.size() == numCourses ? res : vector<int>();
    }
};
// @lc code=end

Analysis

时间复杂度O(N), N为边数.