leetcode#210 Course Schedule

最新推荐文章于 2021-05-17 14:36:15 发布

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本文介绍了一种解决课程先修顺序问题的算法实现，通过拓扑排序确定合理的课程学习顺序。适用于存在先修条件的课程安排场景。

题目：

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].

这道题是一个典型的拓扑排序问题。首先对于所有的course构成的有向图G(V, E)来说，最先选择的课程一定是没有先修课程的course，用图论的语言来表述就是入度为零。这样我们只要维护一个有向图，每次从中删除一个入度为零的点及其连接的边，将其加入到结果中去。当这个图为空时边求出了这个问题的一个解。但是，当这个图不为空又不存在入度为零的点时，说明这个图中存在环，即不存在解。代码如下：

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        int pre[numCourses];
        for (int i=0; i<numCourses; i++) pre[i]=0;
        vector<vector<int>> adj(numCourses);
        for (int i=0; i<prerequisites.size(); i++) {
            pair<int, int>& tmp = prerequisites[i];
            adj[tmp.second].push_back(tmp.first);
            pre[tmp.first]++;
        }
        vector<int> result;
        int count = 0;
        while(count != numCourses) {
            count = 0;
            for(int i=0; i<numCourses; i++) if (pre[i]==0) {
                pre[i]=-1;
                vector<int> tmp = adj[i];
                for(int j=0; j<tmp.size(); j++) pre[tmp[j]]--;
                result.push_back(i);
            } else count++;
        }
        if (result.size()!=numCourses) result ={};
        return result;
    }
};