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].
Note:
The input prerequisites is a graph represented by a list of edges, not adjacency matrices. Read more about how a graph is represented.
[分析] 对Course Schedule的简单扩展,关键点仍然是要考虑prerequisites 中可能有重复,
因此遍历prerequisites时要注意indegree数组不要重复累加,否则会得到错误答案。
ublic class Solution {
public int[] findOrder(int numCourses, int[][] prerequisites) {
int[] result = new int[numCourses];
if (prerequisites == null || prerequisites.length == 0 || prerequisites[0].length == 0) {
for (int i = 0; i < numCourses; i++)
result[i] = i;
return result;
}
//build graph
ArrayList<HashSet<Integer>> graph = new ArrayList<HashSet<Integer>>(numCourses);
int[] indegree = new int[numCourses];
for (int i = 0; i < numCourses; i++)
graph.add(new HashSet<Integer>());
int edges = prerequisites.length;
for (int i = 0; i < edges;i++) {
if (graph.get(prerequisites[i][1]).add(prerequisites[i][0])) //考虑prerequisites可能有重复
indegree[prerequisites[i][0]]++;
}
//search
LinkedList<Integer> queue = new LinkedList<Integer>();
for (int i = 0; i < numCourses; i++) {
if (indegree[i] == 0)
queue.offer(i);
}
int counter = 0;
while (!queue.isEmpty()) {
int course = queue.poll();
result[counter] = course;
counter++;
for (int w : graph.get(course)) {
if (--indegree[w] == 0) queue.offer(w);
}
}
if (counter == numCourses)
return result;
else
return new int[0];
}
}
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