[LeetCode][JavaScript]Course Schedule

Course Schedule

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.

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 if a cycle exists 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.

https://leetcode.com/problems/course-schedule/

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选课，[m,n]，选m的先决条件是要先选n。

拓扑排序，用DFS实现。

用邻接表建图。循环遍历建立所有节点，再循环所有的边把图建出来。

最后遍历所有的点进行DFS。

两个关键的哈希表：visited和visitedPath。

visited是DFS外层循环用的，访问过的点就不再访问了，节省时间；visitedPath是记录回溯的路径，这东西是本题的关键。

举例来说canFinish(4, [[1,0],[2,0],[3,1],[3,2]]) ：

             

走0 -> 1 -> 3这条路的时候，会访问到3，visited[3]记成true了。

当走0 -> 2 -> 3的时候，如果只是看visited变量，3已经访问过了。

所以需要visitedPath记录当前路径上的点，当一条路走不通回溯的时候，visitedPath也需要回溯。

/**
 * @param {number} numCourses
 * @param {number[][]} prerequisites
 * @return {boolean}
 */
var canFinish = function(numCourses, prerequisites) {
    var map = {}, visited = {}, visitedPath = {};
    //build map
    for(i = 0; i < numCourses; i++){
        map[i] = { index : i, next : [] };
    }
    //add edge
    for(i = 0; i < prerequisites.length; i++){
        map[prerequisites[i][1]].next.push(map[prerequisites[i][0]]);
    }
    //bfs
    for(i = 0; i < numCourses; i++){
        if(!visited[i]){
            if(!bfs(map[i])){
                return false;
            }    
        }        
    }
    return true;

    function bfs(node){
        visited[node.index] = true;
        visitedPath[node.index] = true;
        if(node.next.length > 0){
            for(var i = 0; i < node.next.length; i++){
                if(visitedPath[node.next[i].index]){
                    return false;
                } 
                if(!bfs(map[node.next[i].index])){
                    return false;
                }
            }
        }
        visitedPath[node.index] = false;
        return true;
    }
};