[LeetCode] 207. Course Schedule 课程安排

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.

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.

给定n个课程，上课的顺序有先后要求，用pair表示，判断是否能完成所有课程。

对于每一对课程的顺序关系，把它看做是一个有向边，边是由两个端点组成的，用两个点来表示边，所有的课程关系即构成一个有向图，问题相当于判断有向图中是否有环。判断有向图是否有环的方法是拓扑排序。

拓扑排序：维护一张表记录所有点的入度，移出入度为0的点并更新其他点的入度，重复此过程直到没有点的入度为0。如果原有向图有环的话，此时会有剩余的点且其入度不为0；否则没有剩余的点。

图的拓扑排序可以DFS或者BFS。遍历所有边，计算点的入度；将入度为0的点移出点集，并更新剩余点的入度；重复步骤2，直至没有剩余点或剩余点的入度均大于0。

这里不能使用邻接矩阵，应该使用邻接表来存储有向图的信息。邻接表可以使用结构体来实现，每个结构体存储一个值以及一个指向下一个节点的指针，同时维护一个存储多个头结点的数组即可。除此之外，在数据结构简单的情况下，还可以使用数组来模拟简单的邻接表。

Java：BFS

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        int[] pre = new int[numCourses];
        List<Integer>[] satisfies = new List[numCourses];
        for(int i=0; i<numCourses; i++) satisfies[i] = new ArrayList<>();
        for(int i=0; i<prerequisites.length; i++) {
            satisfies[prerequisites[i][1]].add(prerequisites[i][0]);
            pre[prerequisites[i][0]] ++;
        }
        int finish = 0;
        LinkedList<Integer> queue = new LinkedList<>();
        for(int i=0; i<numCourses; i++) {
            if (pre[i] == 0) queue.add(i);
        }
        while (!queue.isEmpty()) {
            int course = queue.remove();
            finish ++;
            if (satisfies[course] == null) continue;
            for(int c: satisfies[course]) {
                pre[c] --;
                if (pre[c] == 0) queue.add(c);
            }
        }
        return finish == numCourses;
    }
}

Java：DFS

public class Solution {
    private boolean[] canFinish;
    private boolean[] visited;
    private List<Integer>[] depends;
    private boolean canFinish(int course) {
        if (visited[course]) return canFinish[course];
        visited[course] = true;
        for(int c: depends[course]) {
            if (!canFinish(c)) return false;
        }
        canFinish[course] = true;
        return canFinish[course];
    }
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        canFinish = new boolean[numCourses];
        visited = new boolean[numCourses];
        depends = new List[numCourses];
        for(int i=0; i<numCourses; i++) depends[i] = new ArrayList<Integer>();
        for(int i=0; i<prerequisites.length; i++) {
            depends[prerequisites[i][0]].add(prerequisites[i][1]);
        }
        for(int i=0; i<numCourses; i++) {
            if (!canFinish(i)) return false;
        }
        return true;
    }
}

Python：

import collections

class Solution(object):
    def canFinish(self, numCourses, prerequisites):
        """
        :type numCourses: int
        :type prerequisites: List[List[int]]
        :rtype: bool
        """
        zero_in_degree_queue, in_degree, out_degree = collections.deque(), {}, {}

        for i, j in prerequisites:
            if i not in in_degree:
                in_degree[i] = set()
            if j not in out_degree:
                out_degree[j] = set()
            in_degree[i].add(j)
            out_degree[j].add(i)

        for i in xrange(numCourses):
            if i not in in_degree:
                zero_in_degree_queue.append(i)

        while zero_in_degree_queue:
            prerequisite = zero_in_degree_queue.popleft()

            if prerequisite in out_degree:
                for course in out_degree[prerequisite]:
                    in_degree[course].discard(prerequisite)
                    if not in_degree[course]:
                        zero_in_degree_queue.append(course)

                del out_degree[prerequisite]

        if out_degree:
            return False

        return True

C++: BFS

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;
    }
};

C++: DFS

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;
    }
};

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