Leetcode 210. Course Schedule II （利用拓扑排序）

最新推荐文章于 2024-01-21 17:17:30 发布

Lin_QC

最新推荐文章于 2024-01-21 17:17:30 发布

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本文深入探讨了一种基于图论的课程排序算法，通过拓扑排序解决课程先修问题，给出实例解析并提供C++实现代码。

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一、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.

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:

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.

二、解题思路

这就是一个图论的问题，我们可以通过拓扑排序来解决。

对于题中的例2

我们可以得到这样一个拓扑图（被指向代表要先上）

我们可以看到0的出度是0，1和2的出度为1，3的出度为2

这时候我们就可以知道，出度为0的要先上，所以先上0

上完0后，把0点去掉，1和2的出度也为0了，这时候可以上1和2

1和2 上完后，3的出度也为0了，上3

所以上课顺序为 0，1，2，3 或0，2，1，3

三、代码

class Solution {
public:
    vector<int> findOrder(int numCourses, vector<vector<int>>& prerequisites) {
        queue<int> zeroNode;  
        vector<int> degree(numCourses, 0);
        vector<int> res;
        for(int i=0;i<prerequisites.size();i++){
            degree[prerequisites[i][0]]++;
        }
        for(int i=0;i<numCourses;i++){
            if(degree[i]==0) 
            {
                res.push_back(i);
                zeroNode.push(i);
            }
        }
        while(!zeroNode.empty()){
            int node = zeroNode.front();
            zeroNode.pop();
            for(int i=0;i<prerequisites.size();i++){
                if(prerequisites[i][1]==node){
                    degree[prerequisites[i][0]]--;
                    if(degree[prerequisites[i][0]]==0){
                        res.push_back(prerequisites[i][0]);
                        zeroNode.push(prerequisites[i][0]);
                    }
                }
            }
        }
        if(res.size()!=numCourses) return vector<int>();
        else return res;
        
    }
};