LeetCode 207. Course Schedule 210. Course Schedule II

最新推荐文章于 2021-12-10 13:25:51 发布

i_chase

最新推荐文章于 2021-12-10 13:25:51 发布

阅读量165

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/qq_39638957/article/details/95180707

博客围绕课程先决条件问题展开，有两道题，一是判断能否完成所有课程，二是给出完成课程的顺序。两道题本质是拓扑排序问题，文中给出了bfs解法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

207

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:

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

210

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.

Example 1:

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

Example 2:

Input: 4, [[1,0],[2,0],[3,1],[3,2]]
Output: [0,1,2,3] or [0,2,1,3]
Explanation: 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.
You may assume that there are no duplicate edges in the input prerequisites.

两道题目其实一样的，其实就是拓扑排序。这里给出bfs解法。

207

class Solution {
    
    
    //拓扑排序
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        int[] incDegrees = new int[numCourses];
        List<List<Integer>> adjs = new ArrayList<>();
        
        for(int i = 0;i<numCourses;i++){
            adjs.add(new ArrayList<>());
        }
        for(int[] edge : prerequisites){
            incDegrees[edge[0]]++;
            adjs.get(edge[1]).add(edge[0]);
        }
        
        //bfs
        Queue<Integer> tovisit = new LinkedList<>();
        int[] res = new int[numCourses];
        int visited = 0;
        for(int i = 0;i<numCourses;i++){
            if(incDegrees[i]==0){
                tovisit.offer(i);
            }
        }
        while(!tovisit.isEmpty()){
            int from = tovisit.poll();
            res[visited] = from;
            visited++;
            for(int to:adjs.get(from)){
                incDegrees[to]--;
                if(incDegrees[to]==0){
                    tovisit.offer(to);
                }
            }
        }
        return visited==numCourses;
    }
    
}

210

class Solution {
    
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        int[] incDegrees = new int[numCourses];
        List<List<Integer>> adjs = new ArrayList<>();
        
        for(int i = 0;i<numCourses;i++){
            adjs.add(new ArrayList<>());
        }
        for(int[] edge : prerequisites){
            incDegrees[edge[0]]++;
            adjs.get(edge[1]).add(edge[0]);
        }
        
        //bfs
        Queue<Integer> tovisit = new LinkedList<>();
        int[] res = new int[numCourses];
        int visited = 0;
        for(int i = 0;i<numCourses;i++){
            if(incDegrees[i]==0){
                tovisit.offer(i);
            }
        }
        while(!tovisit.isEmpty()){
            int from = tovisit.poll();
            res[visited] = from;
            visited++;
            for(int to:adjs.get(from)){
                incDegrees[to]--;
                if(incDegrees[to]==0){
                    tovisit.offer(to);
                }
            }
        }
        if(visited!=numCourses) return new int[0];
        return res;
    }
    
}