207. Course Schedule

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.

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.

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. Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
3. Topological sort could also be done via BFS.

首先用的是拓扑消除法。用matrix保存所有出度，indegree保存每个课的入度个数。queue弹进所有入度为0的课。如果queue不为空，弹出一个课，count+1，遍历所有出度，如果为1，相应的入度-1，如果某一个入度为零了，立即入栈。最后返回count == nomCourse，即是否消除完所有的入度。代码如下：

public class Solution {    public boolean canFinish(int numCourses, int[][] prerequisites) {        int[][] matrix = new int[numCourses][numCourses];        int[] indegree = new int[numCourses];        for (int[] pair: prerequisites) {            indegree[pair[0]]++;            matrix[pair[1]][pair[0]] = 1;        }        Queue<Integer> queue = new LinkedList<Integer>();        for (int i = 0; i < numCourses; i ++) {            if (indegree[i] == 0) queue.offer(i);        }        int count = 0;        while (!queue.isEmpty()) {            int course = queue.poll();            count ++;            for (int i = 0; i < numCourses; i ++) {                if (matrix[course][i] == 1) {                    if (--indegree[i] == 0) {                        queue.offer(i);                    }                }            }        }        return count == numCourses;    }}

另一种方法，深度优先遍历。设置marked标记此序列已经被遍历过，如果被标记过，恰好stacked[course]为true的话说明stack了，loop设置为true，最后返回!loop。代码如下：

public class Solution {    boolean[] marked;    boolean[] stacked;    boolean loop;        public boolean canFinish(int numCourses, int[][] prerequisites) {        ArrayList[] lists = new ArrayList[numCourses];        marked = new boolean[numCourses];        stacked = new boolean[numCourses];        loop = false;        for (int i = 0; i < numCourses; i ++) {            lists[i] = new ArrayList<Integer>();        }        for (int[] pair: prerequisites) {            lists[pair[0]].add(pair[1]);        }        for (int i = 0; i < numCourses; i ++) {            if (!marked[i]) {                helper(lists, i);            }        }        return !loop;    }        private void helper(List<Integer>[] lists, int course) {        stacked[course] = true;        marked[course] = true;        if (loop) return;        for (int i = 0; i < lists[course].size(); i ++) {            int index = lists[course].get(i);            if (!marked[index]) {                helper(lists, index);            } else if (stacked[index]) {                loop = true;            }        }        stacked[course] = false;    }}