leetcode [207]Course Schedule

Posted 2021-10-24 xiaobaituyun

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:

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.

题目大意：

其实就是判断一个有向图中是否存在环，可以使用BFS和DFS做，BFS的效率比DFS要高一些

解法：

java:

BFS解法：

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        //记录有向图图节点之间的关系
        ArrayList[] graph=new ArrayList[numCourses];
        //记录每个节点的入度
        int[] degree=new int[numCourses];
        Queue queue=new LinkedList();
        int count=0;
        
        for (int i=0;i<numCourses;i++) graph[i]=new ArrayList();
        
        for (int i=0;i<prerequisites.length;i++){
            degree[prerequisites[i][1]]++;
            graph[prerequisites[i][0]].add(prerequisites[i][1]);
        }
        
        for (int i=0;i<numCourses;i++){
            if(degree[i]==0){
                ((LinkedList) queue).add(i);
                count++;
            }
        }
        
        while (!queue.isEmpty()){
            int course=(int)((LinkedList) queue).pollFirst();
            for (int i=0;i<graph[course].size();i++){
                int index=(int)graph[course].get(i);
                degree[index]--;
                if (degree[index]==0){
                    ((LinkedList) queue).add(index);
                    count++;
                }
            }
        }
        
        if(count==numCourses) return true;
        else return false;
    }
}

DFS解法;

class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        //记录有向图图节点之间的关系
        ArrayList[] graph=new ArrayList[numCourses];
        //记录每个节点是否被访问过
        boolean[] visited=new boolean[numCourses];

        for (int i=0;i<numCourses;i++) graph[i]=new ArrayList();

        for (int i=0;i<prerequisites.length;i++){
            graph[prerequisites[i][0]].add(prerequisites[i][1]);
        }

        for (int i=0;i<numCourses;i++){
            if (!dfs(graph,visited,i)) return false;
        }

        return true;
    }

    private boolean dfs(ArrayList[] graph, boolean[] visited, int course){
        if(visited[course])
            return false;
        else
            visited[course] = true;;

        for(int i=0; i<graph[course].size();i++){
            if(!dfs(graph,visited,(int)graph[course].get(i)))
                return false;
        }
        visited[course] = false;
        return true;
    }
}