題目連接
https://leetcode.com/problems/course-schedule-ii/??
Course Schedule II
Description
There are a total of n courses you have to take, labeled from 0 to n - 1.
leetcode 課程表？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.
LEETCODE、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$[0,2,1,3].$
LeetCode Online Judge。拓撲排序，輸出結果
Ps:我喜歡用鏈式前向星建圖。
class Solution {
public:vector<int> findOrder(int numCourses, vector<pair<int, int>>& prerequisites) {ans.clear();int m = prerequisites.size();inq = new int[numCourses + 10];memset(inq, 0, sizeof(int)* (numCourses + 10));head = new int[numCourses + 10];memset(head, -1, sizeof(int)* (numCourses + 10));G = new edge[m + 10];for (int i = 0; i < m; i++) {int u = prerequisites[i].first, v = prerequisites[i].second;inq[u]++;add_edge(v, u);}queue<int> q;for (int i = 0; i < numCourses; i++) { if (!inq[i]) q.push(i); }while (!q.empty()) {int u = q.front(); q.pop();ans.push_back(u);for (int i = head[u]; ~i; i = G[i].next) {if (--inq[G[i].to] == 0) q.push(G[i].to);}}delete[]G; delete[]inq, delete[]head;return ans.size() == numCourses ? ans : vector<int>();}
private:vector<int> ans;int tot, *inq, *head;struct edge { int to, next; }*G;inline void add_edge(int u, int v) {G[tot].to = v, G[tot].next = head[u], head[u] = tot++;}
};
