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.

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

Solution：拓扑排序

思路：
Time Complexity: O(E+V) Space Complexity: O(E+V)

Solution Code：

public class Solution {
    public boolean sequenceReconstruction(int[] org, List<List<Integer>> seqs) {
        
        // build graph
        
        Map<Integer, Set<Integer>> graph = new HashMap<>();
        Map<Integer, Integer> indegree = new HashMap<>();
        
        for(List<Integer> seq: seqs) {
            if(seq.size() == 1) {
                if(!graph.containsKey(seq.get(0))) {
                    graph.put(seq.get(0), new HashSet<>());
                    indegree.put(seq.get(0), 0);
                }
            } else {
                for(int i = 0; i < seq.size() - 1; i++) {
                    if(!graph.containsKey(seq.get(i))) {
                        graph.put(seq.get(i), new HashSet<>());
                        indegree.put(seq.get(i), 0);
                    }

                    if(!graph.containsKey(seq.get(i + 1))) {
                        graph.put(seq.get(i + 1), new HashSet<>());
                        indegree.put(seq.get(i + 1), 0);
                    }

                    /*
                    if(!graph.get(seq.get(i)).contains(seq.get(i + 1))) {
                        graph.get(seq.get(i)).add(seq.get(i + 1));
                        indegree.put(seq.get(i + 1), indegree.get(seq.get(i + 1)) + 1);
                    }
                    */
                    if(graph.get(seq.get(i)).add(seq.get(i + 1))) {
                        indegree.put(seq.get(i + 1), indegree.get(seq.get(i + 1)) + 1);
                    }
                }
            }
        }

        Queue<Integer> queue = new LinkedList<>();
        for(Map.Entry<Integer, Integer> entry: indegree.entrySet()) {
            if(entry.getValue() == 0) queue.offer(entry.getKey());
        }

        int index = 0;
        while(!queue.isEmpty()) {
            int queue_size = queue.size();
            if(queue_size > 1) return false; // 关键
            int curr = queue.poll();
            if(index == org.length || curr != org[index++]) return false; //与给的对比
            for(int next: graph.get(curr)) {
                indegree.put(next, indegree.get(next) - 1);
                if(indegree.get(next) == 0) queue.offer(next);
            }
        }
        return index == org.length && index == graph.size();
    }
}