Leetcode210 Course ScheduleII

题目描述

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] .

思路

DFS。两个表，正在访问visiting和已访问过visited。每一门课与他的前置课连接，形成graph。如果访问过则返回true，直到形成闭环，再将节点变为未访问过。

代码

class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        ArrayList<ArrayList<Integer>> graph = new ArrayList<>();
        for(int i=0;i<numCourses;i++){
            graph.add(new ArrayList<Integer>());
        }
        for(int i=0;i<prerequisites.length;i++){
            int course = prerequisites[i][0];
            int prerequisite =prerequisites[i][1];
            graph.get(course).add(prerequisite);
        }
        int[] visited = new int[numCourses];
        List<Integer> ans = new ArrayList<>();
        Integer index = numCourses;
        for(int i=0;i<numCourses;i++){
            if(dfs(i,graph,visited,ans))
                return new int[0];
        }
        int[] array = new int[ans.size()];
        for(int i = 0; i < ans.size();i++){
            array[i] = ans.get(i);
        }
        return array;
    }
    
    private boolean dfs(int cur, ArrayList<ArrayList<Integer>> graph,int[] visited, List<Integer> ans){
        if(visited[cur]==1) return true;
        if(visited[cur]==2) return false;
        visited[cur] = 1;
        for(int i: graph.get(cur))
            if(dfs(i,graph,visited,ans))
                return true;
        visited[cur]=2;
        ans.add(cur);
        return false;
    }
}