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[LeetCode] Course Schedule 课程清单

时间:2015-05-07 14:26:31      阅读:167      评论:0      收藏:0      [点我收藏+]

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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.

click to show more hints.

Hints:
  1. 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.
  2. 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?
  3. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
  4. 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数啊,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;
    }
};

 

 

 

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] Course Schedule 课程清单

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原文地址:http://www.cnblogs.com/grandyang/p/4484571.html

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