算法设计作业4

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本文针对课程安排问题，提出一种基于拓扑排序的解决方案。通过跟踪课程先决条件，实现合理安排学习顺序，确保所有课程都能顺利完成。文章详细介绍了算法的设计思路及实现过程。

第四周作业：

210.Course Schedule II

解题思路：

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.

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].

思路：跟上一周的Course Schedule之前的要求是一样的，完成一个拓扑排序，只不过附加条件是并同时输出排序结果。所以其实难度也不是很大。依次把入度为0的点加入一个序列中即可。

代码如下：

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {
        vector<int> result; 
        vector<int> resultnull;
        
        int * in_deg;  
        in_deg=new int[numCourses];  
        for(int i=0;i<numCourses;i++)  
          in_deg[i]=0;  
        for(int i=0;i<prerequisites.size();i++)  
            in_deg[prerequisites[i].first]++;  
        queue<int> q;  
        for(int i=0;i<numCourses;i++)  
            if(in_deg[i]==0)  
                 q.push(i);  
          
        int count=q.size();  
        while(!q.empty()){  
            int temp=q.front(); 
            result.push_back(temp);
            q.pop();  
            for(int i=0;i<prerequisites.size();i++){  
                if(temp==prerequisites[i].second){  
                     in_deg[prerequisites[i].first]--;  
                     if(in_deg[prerequisites[i].first]==0){  
                         q.push(prerequisites[i].first);  
                         count++;  
                     }  
                }  
            }  
        }  
        if(count == numCourses)
        return result;
        else
        return resultnull;  
    }
};