LeetCode-Course Schedule II

Course Schedule II

##题目

####Course Schedule II

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.

click to show more hints.

#######Hints:

  • This problem is equivalent to finding the topological order in a directed graph. If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses.
  • Topological Sort via DFS - A great video tutorial (21 minutes) on Coursera explaining the basic concepts of Topological Sort.
  • Topological sort could also be done via BFS.

##解题思路
该题的思路与Course Schedule类似，只不过在这题中需要保存最后拓扑结构的序列，因此在返回结果的时候稍微有所变动。

##算法代码
代码采用JAVA实现：

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public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {
        Graph graph=new Graph(numCourses,prerequisites);
        return graph.topoGraph();
    }

    class Graph{
    	private int n; // 图中节点数量
    	private HashMap<Integer,ArrayList<Integer>> alledges; //所有的边
    	private int[] indegree; //每个节点的入度
    	private int[] outdegree; //每个节点的出度

    	public Graph(int num,int[][] edges)
    	{
    		this.n=num;
    		alledges=new HashMap<Integer,ArrayList<Integer>>();
    		indegree=new int[n];
    		outdegree=new int[n];
    		for(int i=0;i<edges.length;i++){
    			if(alledges.containsKey(edges[i][1])){
    				ArrayList<Integer> tmp=alledges.get(edges[i][1]);
    				tmp.add(edges[i][0]);
    				alledges.put(edges[i][1],tmp);
    			}else{
    				ArrayList<Integer> tmp=new ArrayList<Integer>();
    				tmp.add(edges[i][0]);
    				alledges.put(edges[i][1],tmp);
    			}
    			indegree[edges[i][0]]++;
    			outdegree[edges[i][1]]++;
    		}

    	}

    	public int[] topoGraph()
    	{
    		int count=0;
    		int[] result=new int[n];
    		//对其进行拓扑结构求解
    		//定义个队列其中存放入度为0的节点
    		LinkedList<Integer> queen=new LinkedList<Integer>();
    		for(int i=0;i<indegree.length;i++)
    		{
    			if(indegree[i]==0)
    				queen.offer(i);
    		}

    		while(!queen.isEmpty()){
    			int tmp=queen.poll();
    			result[count]=tmp;
    			count++;
    			//对该节点其所有孩子节点的入度进行-1操作，并判断是否为0
    			ArrayList<Integer> childs=alledges.get(tmp);
    			if(childs!=null){
    				for(int child:childs){
    					indegree[child]--;
    					if(indegree[child]==0)
    						queen.offer(child);
    				}
    			}
    		}

    		if(count==n) //表示完美生成拓扑结构，即可以完成所有课程
    			return result;
    		else
    		{
    			return new int[]{}; //返回一个空的数组
    		}
    	}

    }
}

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