[Leetcode] Course Schedule, Problem solving report

Topics

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.

Ideas

This topic is a typical topology of the graph of the problem, here first briefly introduce the topology sequencing algorithm implementation steps:

1. The initialization graph adjacency matrix (U depends on V, then matrix[u][v]==1) and each node's in-degree group (U depends on V, then indegree[v]++).
2. Select a vertex output without a precursor node (that is, indegree[v]==0) from the into degree group.
3. Delete the vertex and all arcs with it as the head node. (i.e. all matrix[v][otherv]==1, indegre[otherv]–).
4. Repeat the 2,3 step until all vertices have been output, or there are no vertices without precursors in the diagram. (There is a loop in the illustration).

So, the problem is to detect whether there is a loop in the last constructed graph (that is, if the Indegree array has a vertex value greater than 0).

AC Code

The AC code is as follows:

 Public  class solution {     Public Boolean Canfinish(intNumcourses,int[] Prerequisites) {BooleanCanfinish =true;int[] Indegree =New int[Numcourses];int[] Matrix =New int[Numcourses] [Numcourses];//Initial data         for(inti =0; i < prerequisites.length; i + +) {if(matrix[prerequisites[i][0]][prerequisites[i][1]] ==1)Continue; indegree[prerequisites[i][1]] ++; matrix[prerequisites[i][0]][prerequisites[i][1]] =1; }//Topological sort         for(inti =0; i < numcourses; i + +) { for(intj =0; J < Numcourses; J + +) {if(Indegree[j] = =0) {Indegree[j]--;//Delete node which start is J                     for(intK =0; K < numcourses; K + +) {if(Matrix[j][k] = =1) {Indegree[k]--; MATRIX[J][K] =0; }                    } Break; }            }        }//Output result         for(inti =0; i < numcourses; i + +) {if(Indegree[i] >0) {Canfinish =false; Break; }        }returnCanfinish; }}

[Leetcode] Course Schedule, Problem solving report