210. Course Schedule II

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题目：

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.

链接： http://leetcode.com/problems/course-schedule-ii/

题解：

跟Course Schedule一样，这次是求出拓扑排序后的顺序。我们依然是用Kahn‘s Algorithm和Tarjan‘s Algorithm。

Kahn‘s Algorithm:

public class Solution {
    public int[] findOrder(int numCourses, int[][] prerequisites) {     // Kahn‘s Algorithms
        if(numCourses <= 0)
            return new int[]{};
        int[] res = new int[numCourses];
        for(int i = 0; i < numCourses; i++)
            res[i] = i;
            
        if(prerequisites == null || prerequisites.length == 0)
            return res;
        
        int[] inDegree = new int[numCourses];
        for(int[] edge : prerequisites)
            inDegree[edge[1]]++;
        
        Queue<Integer> queue = new LinkedList<>();
        for(int i = 0; i < numCourses; i++)         // calculate inDegree
            if(inDegree[i] == 0)
                queue.offer(i);
        
        int index = numCourses - 1;
        while(!queue.isEmpty()) {
            int source = queue.poll();
            res[index--] = source;              // reverse post order
            for(int[] edge : prerequisites) {
                if(edge[0] == source) {
                    inDegree[edge[1]]--;
                    if(inDegree[edge[1]] == 0)
                        queue.offer(edge[1]);
                }
            }
        }
        
        if(index < 0) {             //looped through all vertex
            return res;
        } else
            return new int[]{};     // has cycle, not DAG
    }
}

Reference:

https://leetcode.com/discuss/49696/easy-understanding-dfs-solution-in-c

https://leetcode.com/discuss/36572/java-code-for-course-schedule-ii

https://leetcode.com/discuss/42710/java-dfs-double-cache-visiting-each-vertex-once-433ms

https://leetcode.com/discuss/35605/two-ac-solution-in-java-using-bfs-and-dfs-with-explanation

https://leetcode.com/discuss/35787/simple-ac-c-dfs-toposort-solution

https://leetcode.com/discuss/35590/dfs-solution-in-java

210. Course Schedule II

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原文地址：http://www.cnblogs.com/yrbbest/p/4977294.html