Check whether the original sequence org can be uniquely reconstructed from the sequences in seqs. The org sequence is a permutation of the integers from 1 to n, with 1 ≤ n ≤ 104. Reconstruction means building a shortest common supersequence of the sequences in seqs (i.e., a shortest sequence so that all sequences in seqs are subsequences of it). Determine whether there is only one sequence that can be reconstructed from seqs and it is the org sequence.

Example 1:

Input:
org: [1,2,3], seqs: [[1,2],[1,3]]

Output:
false

Explanation:
[1,2,3] is not the only one sequence that can be reconstructed, because [1,3,2] is also a valid sequence that can be reconstructed.

Explanation:
[1,2,3] is not the only one sequence that can be reconstructed, because [1,3,2] is also a valid sequence that can be reconstructed.
Example 2:

Input:
org: [1,2,3], seqs: [[1,2]]

Output:
false

Explanation:
The reconstructed sequence can only be [1,2].
Example 3:

Input:
org: [1,2,3], seqs: [[1,2],[1,3],[2,3]]

Output:
true

Explanation:
The sequences [1,2], [1,3], and [2,3] can uniquely reconstruct the original sequence [1,2,3].
Example 4:

Input:
org: [4,1,5,2,6,3], seqs: [[5,2,6,3],[4,1,5,2]]

Output:
true
UPDATE (2017/1/8):
The seqs parameter had been changed to a list of list of strings (instead of a 2d array of strings). Please reload the code definition to get the latest changes.

看的答案，有些细节没看懂

class Solution {
    public boolean sequenceReconstruction(int[] org, List<List<Integer>> seqs) {
        Map<Integer, Set<Integer>> neighbors = new HashMap<>();
        Map<Integer, Integer> indegree = new HashMap<>();
        
        //build a graph according to the given sequences.
        for(int i = 0; i < seqs.size(); i++){
            List<Integer> seq = seqs.get(i);
            if (seq.size() == 1){
                indegree.putIfAbsent(seq.get(0), 0);
                neighbors .putIfAbsent(seq.get(0), new HashSet<>());    
            } else {
                for (int j = 0; j < seq.size() - 1; j++){
                    indegree.putIfAbsent(seq.get(j), 0);
                    indegree.putIfAbsent(seq.get(j + 1), 0);
                    neighbors.putIfAbsent(seq.get(j), new HashSet<>());   
                    neighbors.putIfAbsent(seq.get(j + 1), new HashSet<>()); 
                    //why
                    if (!neighbors.get(seq.get(j)).contains(seq.get(j + 1))){
                        indegree.put(seq.get(j + 1), indegree.get(seq.get(j + 1)) + 1);
                    }
                    neighbors.get(seq.get(j)).add(seq.get(j + 1));
                }
            }
        } 
        boolean start = false;
        int point = 0;
        for (Integer key : indegree.keySet()){
            if (indegree.get(key) == 0){
                start = true;
                point = key;
            }
        }
        if (!start){
            return false;
        }
        Queue<Integer> queue = new LinkedList<>();
        queue.offer(point); 
        int index = 0;
        while (!queue.isEmpty()){
            int size = queue.size();
            if (size > 1){
                return false;
            }
            int num = queue.poll();
            //why
            if (index == org.length || org[index] != num){
                return false;
            }
            index++;
            for (Integer nei : neighbors.get(num)){
                indegree.put(nei, indegree.get(nei) - 1);
                if (indegree.get(nei) == 0){
                    queue.offer(nei);
                }
            }
        }
        return index == org.length && index == indegree.size();
    }
}