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

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.

思路：

這個題是讓我們遍歷所有 seqs 序列來推斷是否可以唯一確定所給出的org，換句話說就是求拓撲排序 ；

這裡對 example 中的例子稍作解釋，例如example 1 與 example 3：

這兩個例子中所給定的 org是完全一樣的[1，2，3]，但是seqs序列中example 1中給定的是[1,2],[2,3]而example 3中給定的是[1,2],[1,3],[2,3]，這裡我們可以假設給定的 seqs 序列是可以邊疊加在同一個有向圖中的依賴序列，那麼可推出：

接著我們對這兩個圖示進行拓撲排序，我們每次都從出度最多的節點出發，在出發前檢查是否這個點是否是隻有出度，如果存在入度，那麼去找它的入度節點中出度數目最大的，那麼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]；所以從這裡我們就可以從example 3中唯一確定原序列，而example 1並不可以。

程式碼：

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;
    }
};