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, is it possible for you to finish all courses?
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 it is possible.
2, [[1,0],[0,1]]
There are a total of 2 courses to take. To take course 1 you should have finished course 0, and to take course 0 you should also have finished course 1. So it is impossible.
- This problem is equivalent to finding if a cycle exists in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
- There are several ways to represent a graph. For example, the input prerequisites is a graph represented by a list of edges. Is this graph representation appropriate?
- Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
- Topological sort could also be done via BFS.
这道课程清单的问题对于我们学生来说应该不陌生，因为我们在选课的时候经常会遇到想选某一门课程，发现选它之前必须先上了哪些课程，这道题给了很多提示，第一条就告诉我们了这道题的本质就是在有向图中检测环。 LeetCode中关于图的题很少，有向图的仅此一道，还有一道关于无向图的题是 Clone Graph 无向图的复制。个人认为图这种数据结构相比于树啊，链表啊什么的要更为复杂一些，尤其是有向图，很麻烦。第二条提示是在讲如何来表示一个有向图，可以用边来表示，边是由两个端点组成的，用两个点来表示边。第三第四条提示揭示了此题有两种解法，DFS和BFS都可以解此题。我们先来看BFS的解法，我们定义二维数组graph来表示这个有向图，一位数组in来表示每个顶点的入度。我们开始先根据输入来建立这个有向图，并将入度数组也初始化好。然后我们定义一个queue变量，将所有入度为0的点放入队列中，然后开始遍历队列，从graph里遍历其连接的点，每到达一个新节点，将其入度减一，如果此时该点入度为0，则放入队列末尾。直到遍历完队列中所有的值，若此时还有节点的入度不为0，则说明环存在，返回false，反之则返回true。代码如下：
解法一：
class Solution { public: bool canFinish(int numCourses, vector<vector<int>>& prerequisites) { vector<vector<int> > graph(numCourses, vector<int>(0)); vector<int> in(numCourses, 0); for (auto a : prerequisites) { graph[a[1]].push_back(a[0]); ++in[a[0]]; } queue<int> q; for (int i = 0; i < numCourses; ++i) { if (in[i] == 0) q.push(i); } while (!q.empty()) { int t = q.front(); q.pop(); for (auto a : graph[t]) { --in[a]; if (in[a] == 0) q.push(a); } } for (int i = 0; i < numCourses; ++i) { if (in[i] != 0) return false; } return true; } };
下面我们来看DFS的解法，也需要建立有向图，还是用二维数组来建立，和BFS不同的是，我们像现在需要一个一维数组visit来记录访问状态，这里有三种状态，0表示还未访问过，1表示已经访问了，-1表示有冲突。大体思路是，先建立好有向图，然后从第一个门课开始，找其可构成哪门课，暂时将当前课程标记为已访问，然后对新得到的课程调用DFS递归，直到出现新的课程已经访问过了，则返回false，没有冲突的话返回true，然后把标记为已访问的课程改为未访问。代码如下：
解法二：
class Solution { public: bool canFinish(int numCourses, vector<vector<int> >& prerequisites) { vector<vector<int> > graph(numCourses, vector<int>(0)); vector<int> visit(numCourses, 0); for (auto a : prerequisites) { graph[a[1]].push_back(a[0]); } for (int i = 0; i < numCourses; ++i) { if (!canFinishDFS(graph, visit, i)) return false; } return true; } bool canFinishDFS(vector<vector<int> > &graph, vector<int> &visit, int i) { if (visit[i] == -1) return false; if (visit[i] == 1) return true; visit[i] = -1; for (auto a : graph[i]) { if (!canFinishDFS(graph, visit, a)) return false; } visit[i] = 1; return true; } };
本文转自博客园Grandyang的博客，原文链接：课程清单[LeetCode] Course Schedule ，如需转载请自行联系原博主。
