LeetCode 210. Course Schedule II 题解

210. Course Schedule II

 

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• Difficulty: Medium

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].

解题思路：

本题即在course Schedule 的基础上进行拓扑排序。首先记录下每个点的入度，然后找到入度为0的顶点，从将该点加入到结果向量中后，删除该点，同时所有该点指向的点的入度减1，继续寻找下一个入度为0 的点，如果没有入度为0的点，并且此时结果中的点小于numCourses,则说明其中有环， 不能进行拓扑排序。

代码展示：

class Solution {public:    int finds(int prere[],int numCourses,bool add[])    {        for(int i=0;i<numCourses;i++)        {            if(prere[i]==0&&add[i]==false)            {                return i;            }        }        return -1;    }    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {        int prere[numCourses]={0};        bool add[numCourses] = {0};        int n = prerequisites.size();        vector<int> g[numCourses];        vector<int> ans;        vector<int> empty;        for(int i=0;i<n;i++)        {            pair<int,int> tmp = prerequisites[i];            prere[tmp.first]++;            g[tmp.first].push_back(tmp.second);        }        for(int i=0;i<numCourses;i++)        {            int n = finds(prere,numCourses,add);            if(n==-1) return empty;            ans.push_back(n);            add[n]=true;            for(int j=0;j<numCourses;j++)            {                for(int k=0;k<g[j].size();k++)                {                    if(g[j][k]==n)                    {                        prere[j]--;                    }                }            }                                }        return ans;                    }};