标签：
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, return the ordering of courses you should take to finish all courses.
There may be multiple correct orders, you just need to return one of them. If it is impossible to finish all courses, return an empty array.
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 the correct course order is [0,1]
4, [[1,0],[2,0],[3,1],[3,2]]
There are a total of 4 courses to take. To take course 3 you should have finished both courses 1 and 2. Both courses 1 and 2 should be taken after you finished course 0. So one correct course order is [0,1,2,3]. Another correct ordering is[0,2,1,3].
解法1：内存超过
public class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
int [][]matrix=new int[numCourses][numCourses];
int [] indegree =new int[numCourses];
int len=prerequisites.length;
for(int i=0;i<len;i++){
int ready=prerequisites[i][0];
int pre=prerequisites[i][1];
if (matrix[pre][ready] == 0)//防止重复条件
indegree[ready]++;
matrix[pre][ready]=1;
}
int count=0;
Queue<Integer> queue =new LinkedList();
for(int i=0;i<indegree.length;i++){
if(indegree[i]==0)
queue.offer(i);
}
int[] ret=new int[numCourses];
while(!queue.isEmpty()){
int course=queue.poll();
ret[count]=course;
count++;
for(int i=0;i<numCourses;i++){
if(matrix[course][i]!=0){
if(--indegree[i]==0){
queue.offer(i);
}
}
}
}
if(count==numCourses)
return ret;
return null;
}
}
运行结果：
方法2：不使用二维矩阵，使用List
代码如下：
public class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
List<Set<Integer>>adjLists=new ArrayList<>();
for(int i=0;i<numCourses;i++){
adjLists.add(new HashSet<Integer>());
}
for(int i=0;i<prerequisites.length;i++){//存的是出度
adjLists.get(prerequisites[i][1]).add(prerequisites[i][0]);
}
int [] indegrees=new int[numCourses];
for(int i=0;i<numCourses;i++){
for(int x:adjLists.get(i)){
indegrees[x]++;
}
}
Queue<Integer>queue=new LinkedList<>();
for(int i=0;i<numCourses;i++){
if(indegrees[i]==0){
queue.offer(i);
}
}
int[] res=new int[numCourses];
int count=0;
while(!queue.isEmpty()){
int cur=queue.poll();
for(int x:adjLists.get(cur)){
indegrees[x]--;
if(indegrees[x]==0){
queue.offer(x);
}
}
res[count++]=cur;
}
if(count==numCourses) return res;
return new int[0];
}
}
运行结果：
(medium)LeetCode 210.Course Schedule II
标签：
原文地址：http://www.cnblogs.com/mlz-2019/p/4716433.html
