leetcode Course Schedule II

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].

Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
题目来源：https://leetcode.com/problems/course-schedule-ii/

分析

可以借助上一题《Course Schedule》的代码，在出队列时保存下来就是拓扑排序了。上一题：http://blog.csdn.net/u010902721/article/details/47274605

代码

class Solution {public:    vector<int> findOrder(int numCourses, vector<pair<int, int> >& prerequisites) {        queue<int> help;//存放入度为0的节点        vector<int> courses(numCourses, 0);//存放节点的入度        vector<int> order;        for(auto edge : prerequisites){            courses[edge.first]++;        }        for(int i = 0; i < numCourses; i++){            if(courses[i] == 0)                help.push(i);        }        while(!help.empty()){            int i = help.front();            help.pop();            order.push_back(i);            for(auto edge : prerequisites){                if(edge.second == i){                    courses[edge.first]--;                    if(courses[edge.first] == 0)                        help.push(edge.first);                }            }        }        if(order.size() != numCourses){//如果有环，那就算了，输出null            vector<int> tmp;            return tmp;        }        else//没环，则成功输出拓扑序列            return order;    }};