leetcode 444. Sequence Reconstruction

leetcode 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.

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思路：

这个题是让我们遍历所有 seqs 序列来推断是否可以唯一确定所给出的org，换句话说就是求拓扑排序 ；

这里对 example 中的例子稍作解释，例如example 1 与 example 3：

这两个例子中所给定的 org是完全一样的[1，2，3]，但是seqs序列中example 1中给定的是[1,2],[2,3]而example 3中给定的是[1,2],[1,3],[2,3]，这里我们可以假设给定的 seqs 序列是可以边叠加在同一个有向图中的依赖序列，那么可推出：

接着我们对这两个图示进行拓扑排序，我们每次都从出度最多的节点出发，在出发前检查是否这个点是否是只有出度，如果存在入度，那么去找它的入度节点中出度数目最大的，那么example 1可推出序列：[1,2,3]或者[1,3,2]；example 3可以推出序列：[1,2,3]；所以从这里我们就可以从example 3中唯一确定原序列，而example 1并不可以。

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代码：

class Solution {
public:
    bool sequenceReconstruction(vector<int>& org, vector<vector<int>>& seqs) {
        if (seqs.empty()) return false;
        int n = org.size(), cnt = n - 1;
        vector<int> pos(n + 1, 0), flags(n + 1, 0);
        bool existed = false;
        for (int i = 0; i < n; ++i) pos[org[i]] = i;
        for (auto& seq : seqs) {
            for (int i = 0; i < seq.size(); ++i) {
                existed = true;
                if (seq[i] <= 0 || seq[i] > n) return false;
                if (i == 0) continue;
                int pre = seq[i - 1], cur = seq[i];
                if (pos[pre] >= pos[cur]) return false;
                if (flags[cur] == 0 && pos[pre] + 1 == pos[cur]) {
                    flags[cur] = 1; --cnt;
                }
            }
        }
        return cnt == 0 && existed;
    }
};