[LeetCode] Course Schedule 课程清单
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<)); vector<); for (auto a : prerequisites) { graph[a[]].push_back(a[]); ++]]; } queue<int> q; ; i < numCourses; ++i) { ) q.push(i); } while (!q.empty()) { int t = q.front(); q.pop(); for (auto a : graph[t]) { --in[a]; ) q.push(a); } } ; i < numCourses; ++i) { ) 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,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<)); vector<); for (auto a : prerequisites) { graph[a[]].push_back(a[]); } ; i < numCourses; ++i) { if (!canFinishDFS(graph, visit, i)) return false; } return true; } bool canFinishDFS(vector<vector<int> > &graph, vector<int> &visit, int i) { ) return false; ) return true; visit[i] = -; for (auto a : graph[i]) { if (!canFinishDFS(graph, visit, a)) return false; } visit[i] = ; return true; } };
类似题目:
参考资料:
http://www.cnblogs.com/easonliu/p/4483437.html
https://leetcode.com/discuss/34741/python-20-lines-dfs-solution-sharing-with-explanation
LeetCode All in One 题目讲解汇总(持续更新中...)
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