[leetcode] 210. Course Schedule II 解题报告

题目链接：https://leetcode.com/problems/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：拓扑排序的应用．拓扑顺序就像是在一个工程中每个小项目完成有向后顺序一样，而我们的任务就是找出这个完成的先后顺利的序列．

其方法就是每次将一个入度为０的结点删除，并且将其指向的结点入度减一．并且每次将入度为０的结点保存起来，最后将其返回即可．和上题基本一样．

代码如下：

class Solution {
public:
    vector findOrder(int numCourses, vector>& prerequisites) {
        vector result;
        vector> graph(numCourses, unordered_set());
        vector inDegree(numCourses, 0);
        for(auto val: prerequisites)
        {
            if(graph[val.second].count(val.first) == 0)//如果当前先修课还没有被计算
            {
                graph[val.second].insert(val.first);//则将此课加入到先修课结点维护的集合中，代表此先修课完成之后才可以学的课
                inDegree[val.first]++;//此课程的入度＋１，代表有多少先修课
            }
        }
        int j;
        for(int i = 0; i < numCourses; i++)
        {
            for(j = 0; j < numCourses; j++)//寻找入度为０的结点，找到就break循环
                if(inDegree[j] == 0) break;
            if(j == numCourses) return vector();//如果最终没有找到，说明有环
            result.push_back(j);
            inDegree[j]--;//将这个入度为０的结点删去
            for(auto val: graph[j])//将入度为０的结点指向的结点入度都－１
                inDegree[val]--;
        }
        return result;
    }
};