444. Sequence Reconstruction

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

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.

這道題可以用拓撲排序，消除法。也可以想簡單一點，只需要檢查在org裏面的兩個相鄰的元素，在seqs也是相鄰的。當然，一些邊界條件還是需要的。代碼如下：

public class Solution {
    public boolean sequenceReconstruction(int[] org, List<List<Integer>> seqs) {
        if (seqs.size() == 0 || org.length == 0) {
            return false;
        }
        int n = org.length;
        int[] pos = new int[n + 1];
        for (int i = 0; i < n; i ++) {
            pos[org[i]] = i;
        }
        boolean[] valid = new boolean[n + 1];
        int totalMatch = n - 1;
        boolean notEmpty = false;
        for (List<Integer> seq: seqs) {
            int size = seq.size();
            if (size > 0) notEmpty = true;
            for (int i = 0; i < size; i ++) {
                if (seq.get(i) < 1 || seq.get(i) > n) return false;
                if (i == 0) continue;
                int x = seq.get(i - 1), y = seq.get(i);
                if (pos[x] >= pos[y]) return false;
                if (valid[x] == false && pos[x] + 1 == pos[y]) {
                    valid[x] = true;
                    totalMatch --;
                }
            }
        }
        return totalMatch == 0 && notEmpty;
    }
}