leetcode 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].

每次取得入度为0的一个节点，访问后更新其他节点

class Solution {//取得入度为零的一门课 int get_one(map<int, vector<int>>&prelist,vector<vector<int>>&adjlist){int ret=-1;for (map<int, vector<int>>::iterator it = prelist.begin(); it != prelist.end();it++){if (it->second.empty()){ret = it->first;for (int i = 0; i < adjlist[it->first].size(); i++)prelist[adjlist[it->first][i]].erase(find(prelist[adjlist[it->first][i]].begin(), prelist[adjlist[it->first][i]].end(), it->first));prelist.erase(it);break;}}return ret;}public:vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {vector<int>re;map<int, vector<int>>prelist;for (int i = 0; i < numCourses; i++)prelist[i] = re;vector<vector<int>> adjlist(numCourses);for (int i = 0; i < prerequisites.size(); i++){prelist[prerequisites[i].first].push_back(prerequisites[i].second);adjlist[prerequisites[i].second].push_back(prerequisites[i].first);}int k = numCourses;while (k > 0){k--;int ret = get_one(prelist, adjlist);if (ret < 0){re.clear(); return re;}re.push_back(ret);}return re;}};

accepted