leetcode-207. Course Schedule

原創

2020-02-20 17:10

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.

Hints:

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.
Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
Topological sort could also be done via BFS.

題意解析：

這道題就是說有一堆課程，有的課程有先修課程，比如說修A之前要修B，然後讓你判斷，能不能修完這些課程。下面的note是說，給你的先修課程是以邊列表呈現的，而不是鄰接矩陣，我們可以注意到題目給我們的是一個二維數組，也就是說[[1,0]]，表示的是修1之前要修0。

這道題是典型的拓撲排序。原理也很簡單，在一個有向圖中，每次找到一個沒有前驅節點的節點（也就是入度爲0的節點），然後把它指向其他節點的邊都去掉，重複這個過程（BFS），直到所有節點已被找到，或者沒有符合條件的節點（如果圖中有環存在）。這些下面的提示其實也告訴你了。DFS或者BFS都可以。DFS代碼如下：

public class Solution {
    public boolean canFinish(int numCourses, int[][] prereq) {
        int[] visited = new int[numCourses];
        
        //courses that would have the dependency graph
        List<List<Integer>> courses = new ArrayList<List<Integer>>();
        
        for(int i=0;i<numCourses;i++){
            courses.add(new ArrayList<Integer>());
        }
        
        //add dependencies
        for(int i=0;i< prereq.length;i++){
            // for a course, add all the courses that are dependent on it
            courses.get(prereq[i][1]).add(prereq[i][0]);
        }
        
        for(int i=0;i<numCourses;i++){
            if(visited[i]==0){
                if(!dfs(i,courses,visited)) return false;
            }
        }
        return true;

    }
    
    public boolean dfs(int i,List<List<Integer>> courses, int[]visited){
        visited[i] = 1;
        
        // these are all the courses which are eligible
        //when we take their prerequisite, which is course i
        List<Integer> eligibleCourses = courses.get(i);
        
        for(int j=0;j<eligibleCourses.size();j++){
            int eligibleCourse = eligibleCourses.get(j);
            //it is already visited, during previous dfs call, so its a cycle 
            if(visited[eligibleCourse] == 1) return false;
            if(visited[eligibleCourse] == 2) continue;
               if(!dfs(eligibleCourse,courses,visited)){
                    return false;
                } 
        }
        
        //it is totally complete and no cycle found in its depth first traversal
        visited[i] = 2;
        return true;
        
        
    }
}

這段代碼並不是我寫的，而是discuss上找到的一個非常高效的算法。我自己寫的BFS只打敗了70，這個打敗了95，所以就不獻醜了。