Course schedule II

Source: Internet
Author: User


There are a total of N courses you have to take, labeled from0Ton - 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 shoshould 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 shoshould 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 shoshould have finished both courses 1 and 2. both courses 1 and 2 shoshould 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].

Solution: Use Topology Sorting, use an array to save the number of pre-courses for each course, and use an array to save the number of post-courses for each subject, then, find the course outputs with the number of pre-courses being 0, and reduce the number of pre-courses for all the post-courses by 1.

The AC code is as follows (Time: 144 ms)

class Solution {public:     vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {vector<int> ans;if (numCourses == 0) return ans;vector<int> preCourseCount(numCourses, 0);vector<bool> processed(numCourses, false);vector<vector<int>> postCourseList(numCourses);int n = prerequisites.size();if (n == 0){for (int i = 0; i < numCourses; ++i){ans.push_back(i);}return ans;}for (int i = 0; i < n; ++i){int post = prerequisites[i].first;int pre = prerequisites[i].second;preCourseCount[post]++;postCourseList[pre].push_back(post);}while (1){if (ans.size() == numCourses) break;int i = 0;for (; i < numCourses; ++i){if (preCourseCount[i] == 0 && !processed[i]){processed[i] = true;ans.push_back(i);vector<int> temp = postCourseList[i];for (int j = 0; j < temp.size(); ++j){preCourseCount[temp[j]]--;}break;}}if (i >= numCourses) return vector<int>(0);}return ans;}};

Course schedule II

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