Leetcode 207. Course Schedule Lesson Plan----------Java

There is a total of n courses you have to take, labeled from 0 to n - 1 .

Some courses May has 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, are it possible for your to finish all courses?

For example:

2, [[1,0]]

There is a total of 2 courses to take. To take course 1 should has finished course 0. So it is possible.

2, [[1,0],[0,1]]

There is a total of 2 courses to take. To take course 1 should has finished course 0, and to take course 0 you should also has 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 on how a graph is represented.
2. Assume that there is no duplicate edges in the input prerequisites.
   
   

The main investigation of the graph of the traversal.

I wrote the time-out, stating that there are still a lot of problems.

1, the Brute force algorithm, the result timeout.

 Public classSolution { Public BooleanCanfinish (intNumcourses,int[] Prerequisites) {                int[] num =New int[numcourses][numcourses];  for(inti = 0; i < prerequisites.length; i++) {num[prerequisites[i][0]][PREREQUISITES[I][1]] = 1; }        int[] finish =New int[numcourses]; BooleanFlag =false;  while(true) {flag=false;  for(inti = 0; i < numcourses; i++){                 for(intj = 0; J < Numcourses; J + +){                    if(Num[i][j] = = 1){                         Break; } Else if(j = = NumCourses-1 && finish[i] = = 0) {Finish[i]= 1;                        Isfinish (num, i); Flag=true; }                }            }            if(Flag = =false){                 for(inti = 0; i < numcourses; i++){                    if(Finish[i] = = 0)                         Break; Else if(i = = NumCourses-1)                        return true; }                return false; }        }    }     Public voidIsfinish (int[] num,intPOS) {         for(inti = 0; i < num.length; i++) {Num[i][pos]= 0; }    }}

2, optimization, but still time-out.

 Public classSolution { Public BooleanCanfinish (intNumcourses,int[] Prerequisites) {                int[] num =New int[numcourses][numcourses];  for(inti = 0; i < prerequisites.length; i++){            if(Num[prerequisites[i][1]][prerequisites[i][0]] = = 1){                return false; } num[prerequisites[i][0]][PREREQUISITES[I][1]] = 1;  for(intj = 0; J < Numcourses; J + +){                /*k = prerequisites[i][1] The work that needs to be done before (i.e. num[k][j] = = 1) needs to be completed num[num[prerequisites[i][0]][                j] = 1; */                if(Num[prerequisites[i][1]][j] = = 1) {num[prerequisites[i][0]][J] = 1; }                /*need to complete k = prerequisites[i][0] to complete the work (num[j][k] = = 1), also to complete num[j][prerequisites[i][1]] = 1; */                if(Num[j][prerequisites[i][0]] = = 1) {num[j][prerequisites[i][1]] = 1; }            }                    }        return true; }    }

3. DFS (refer to discuss)

 Public classSolution { Public BooleanCanfinish (intNumcourses,int[] Prerequisites) {arraylist[] list=Newarraylist[numcourses];  for(inti = 0; i < numcourses; i++) {List[i]=NewArraylist<integer>(); }         for(inti = 0; i < prerequisites.length; i++) {list[prerequisites[i][1]].add (prerequisites[i][0]); }        Boolean[] Visit =New Boolean[numcourses];  for(inti = 0; i < numcourses; i++){            if(!DFS (list, visit, i)) {                return false; }        }        return true; }         Public BooleanDFS (arraylist[) list,Boolean[] Visit,intPOS) {        if(Visit[pos]) {return false; } Else{Visit[pos]=true; }         for(inti = 0; I < list[pos].size (); i++){            if(!dfs (list, visit, (int) (List[pos].get (i) )) {return false; }
List[pos].remove (i); } Visit[pos]=false; return true; }    }

4. BFS

 Public classSolution { Public BooleanCanfinish (intNumcourses,int[] Prerequisites) {List<integer>[] Adj =Newlist[numcourses];  for(inti = 0; i < numcourses; i++) Adj[i]=NewArraylist<integer>(); int[] Indegree =New int[numcourses]; Queue<Integer> readycourses =NewLinkedList (); intFinishcount = 0;  for(inti = 0; i < prerequisites.length; i++)          {            intCurcourse = prerequisites[i][0]; intPrecourse = prerequisites[i][1];            Adj[precourse].add (Curcourse); Indegree[curcourse]++; }         for(inti = 0; i < numcourses; i++)         {            if(Indegree[i] = = 0) Readycourses.offer (i); }         while(!Readycourses.isempty ()) {            intCourse = Readycourses.poll ();//Finishfinishcount++;  for(intNextcourse:adj[course]) {Indegree[nextcourse]--; if(Indegree[nextcourse] = = 0) Readycourses.offer (nextcourse); // Ready            }        }        returnFinishcount = =numcourses; }}

Leetcode 207. Course Schedule Lesson Plan----------Java