LeetCode 210. Course Schedule II（课程安排）

原题网址：https://leetcode.com/problems/course-schedule-ii/

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, return the ordering of courses you should take to finish all courses.

There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.

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 the correct course order is [0,1]

4, [[1,0],[2,0],[3,1],[3,2]]

There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].

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 the topological order 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 int[] findOrder(int numCourses, int[][] prerequisites) {        List<Integer>[] satisfies = new List[numCourses];        int[] requires = new int[numCourses];        for(int i=0; i<numCourses; i++) satisfies[i] = new ArrayList<Integer>();        for(int i=0; i<prerequisites.length; i++) {            requires[prerequisites[i][0]] ++;            satisfies[prerequisites[i][1]].add(prerequisites[i][0]);        }        LinkedList<Integer> queue = new LinkedList<>();        for(int i=0; i<numCourses; i++) {            if (requires[i]==0) queue.add(i);        }        int[] courses = new int[numCourses];        int pos = 0;        while (!queue.isEmpty()) {            int course = queue.remove();            courses[pos++] = course;            for(int next: satisfies[course]) {                requires[next] --;                if (requires[next] == 0) queue.add(next);            }        }        if (pos == courses.length) return courses; else return new int[0];    }}

方法二：广度优先搜索可以再优化如下。

public class Solution {    public int[] findOrder(int numCourses, int[][] prerequisites) {        int[] requires = new int[numCourses];        List<Integer>[] satisfies = new List[numCourses];        for(int i=0; i<numCourses; i++) satisfies[i] = new ArrayList<>();        for(int[] pre: prerequisites) {            requires[pre[0]] ++;            satisfies[pre[1]].add(pre[0]);        }        int[] courses = new int[numCourses];        int course = 0;        for(int i=0; i<numCourses; i++) if (requires[i] == 0) courses[course++] = i;        for(int current = 0; current < numCourses; current ++) {            if (current >= course) return new int[0];            for(int satisfied: satisfies[courses[current]]) {                requires[satisfied] --;                if (requires[satisfied] == 0) courses[course++] = satisfied;            }        }        return courses;    }}

方法三：深度优先搜索。

public class Solution {    private List<Integer>[] depends;    private boolean[] visited;    private boolean[] satisfied;    private int[] courses;    private int pos;    private boolean find(int course) {        if (visited[course]) return satisfied[course];        visited[course] = true;        if (depends[course].size() == 0) {            courses[pos++] = course;            satisfied[course] = true;            return true;        }        for(int depend: depends[course]) if (!find(depend)) return false;        courses[pos++] = course;        satisfied[course] = true;        return true;    }    public int[] findOrder(int numCourses, int[][] prerequisites) {        visited = new boolean[numCourses];        courses = new int[numCourses];        depends = new List[numCourses];        satisfied = new boolean[numCourses];        for(int i=0; i<numCourses; i++) depends[i] = new ArrayList<>();        for(int[] pre: prerequisites) depends[pre[0]].add(pre[1]);        for(int i=0; i<numCourses; i++) if (!find(i)) return new int[0];        return courses;    }}