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.
topological sort, in a directed graph, every time found node without pre-node, then remove all its edges. until iterate all nodes or found cycle.
public boolean canFinish(int numCourses, int[][] prerequisites) {
List<HashSet> preCourse = new ArrayList<HashSet>();
for(int i=0;i<numCourses;i++){
preCourse.add(new HashSet<Integer>());
}
for(int i=0;i<prerequisites.length;i++){
preCourse.get(prerequisites[i][1]).add(prerequisites[i][0]);
}
int[] prenums = new int[numCourses];
for(int i=0;i<numCourses;i++){
Set set = preCourse.get(i);
Iterator<Integer> iterator = set.iterator();
while(iterator.hasNext()){
prenums[iterator.next()]++;
}
}
for(int i=0;i<numCourses;i++){
int j=0;
for(;j<numCourses;j++){
if(prenums[j]==0) break;
}
if(j == numCourses) return false;
prenums[j] = -1;
Set set = preCourse.get(j);
Iterator<Integer> iterator = set.iterator();
while(iterator.hasNext()){
prenums[iterator.next()]--;
}
}
return true;
}
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[LeetCode 207] Course Schedule
原文地址:http://blog.csdn.net/sbitswc/article/details/48194505