LeetCode 207. Course Schedule

原题链接在这里：https://leetcode.com/problems/course-schedule/description/

题目：

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.

题解：

topological sort. 每一门课就是一个vertex, 每一个preRequest就是一个edge.

求Course Schedule, 等同问题是有向图检测环. 一个DAG的Topological Order可以有大于1种.

常用的Topological Sorting算法有两种

1. Kahn's Algorithms (wiki)： BFS based， start from with vertices with 0 incoming edge，insert them into list S，at the same time we remove all their outgoing edges，after that find new vertices with 0 incoming edges and go on. 
2. Tarjan's Algorithms (wiki)： DFS based， loop through each node of the graph in an arbitrary order，initiating a depth-first search that terminates when it hits any node that has already been visited since the beginning of the topological sort or the node has no outgoing edges (i.e. a leaf node). 

Time Complexity: O(V + E). Space: O(V).

AC Java:

public class Solution {
    public boolean canFinish(int numCourses, int[][] prerequisites) {
        //BFS based
        if(numCourses <= 0){
            return true;
        }
        
        //用List<List>建立 adjancy list
        List<List<Integer>> fromTo = new ArrayList<List<Integer>>();
        for(int i = 0; i<numCourses; i++){
            fromTo.add(new ArrayList<Integer>());
        }
        for(int [] edge: prerequisites){
            fromTo.get(edge[1]).add(edge[0]);
        }
        
        //计算每门课的prerequisit count
        int [] inDegree = new int[numCourses];
        for(int [] edge : prerequisites){
            inDegree[edge[0]]++;
        }
        
        //把prerequist count为0的课加到queue里面
        List<Integer> res = new ArrayList<Integer>();
        LinkedList<Integer> que = new LinkedList<Integer>();
        for(int i = 0; i<inDegree.length; i++){
            if(inDegree[i] == 0){
                que.add(i);
            }
        }
        
        //从queue里poll出来的课程 去掉他们的outdegree edge
        while(!que.isEmpty()){
            int source = que.poll();
            res.add(source);
            
            for(int destination: fromTo.get(source)){
                inDegree[destination]--;
                //若是该点的prerequist count减一后变成0, 就加到queue中.
                if(inDegree[destination] == 0){
                    que.add(destination);
                }
            }
        }
        return res.size() == numCourses;
    }
}

跟上Course Schedule II, Course Schedule III, Alien Dictionary, Parallel Courses.

reference: http://www.cnblogs.com/yrbbest/p/4493547.html