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C++实现LeetCode(207.课程清单)

 2021-08-06 22:00:16     

[LeetCode] 207. 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

[LeetCode] 207. 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?

Example 1:

Input: 2, [[1,0]]
Output: true
Explanation: There are a total of 2 courses to take.
To take course 1 you should have finished course 0. So it is possible.

Example 2:

Input: 2, [[1,0],[0,1]]
Output: false
Explanation: 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.

Note:

  1. The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
  2. You may assume that there are no duplicate edges in the input prerequisites.

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。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。先来看 BFS 的解法,定义二维数组 graph 来表示这个有向图,一维数组 in 来表示每个顶点的入度。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>());
vector<int> in(numCourses);
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,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>());
vector<int> visit(numCourses);
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;
    }
};

Github 同步地址:

https:https://github.com/grandyang/leetcode/issues/207

参考资料:

https:https://leetcode.com/problems/course-schedule/

https:https://leetcode.com/problems/course-schedule/discuss/58524/Java-DFS-and-BFS-solution

https:https://leetcode.com/problems/course-schedule/discuss/58516/Easy-BFS-Topological-sort-Java

https:https://leetcode.com/problems/course-schedule/discuss/162743/JavaC%2B%2BPython-BFS-Topological-Sorting-O(N-%2B-E)

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