LeetCode 207. Course Schedule（课程安排）

原题网址：https://leetcode.com/problems/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:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.

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.

方法一：广度优先搜索。

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;    }}

方法二：深度优先搜索。

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;    }}