[Leetcode] 444. Sequence Reconstruction 解题报告

题目：

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.

思路：

一道比较麻烦的拓扑排序题目。

代码：

class Solution {
public:
    bool sequenceReconstruction(vector<int>& org, vector<vector<int>>& seqs) {
        if (seqs.size() == 0) {
            return false;
        } 
        int n = org.size();
        int count = 0;
        unordered_map<int, unordered_set<int>> graph;   // record parents
        vector<int> degree(n + 1, 0);                   // record out degree
        for (auto s : seqs) {
            for (int i = s.size() - 1; i >= 0; --i) {
                if (s[i] > n || s[i] < 0) { 
                    return false;                           // in case number in seqs is out of range 1-n
                }
                if (i > 0 && !graph[s[i]].count(s[i-1])) {  // if not included 
                    graph[s[i]].insert(s[i-1]);
                    if (degree[s[i-1]]++ == 0) 
                        count++;
                }
            }
        }
        if (count < n - 1) {
            return false; // all nodes should have degree larger than 0 except the last one
        }
        for (int i = n-1; i >= 0; --i) {
            if (degree[org[i]] > 0) {
                return false;   // the last node should have 0 degree
            }
            for (auto p : graph[org[i]]) {
                if (--degree[p] == 0 && p != org[i-1]) { // found a node that is not supposed to have 0 degree
                    return false;
                }
            }
        }
        return true;
    }
};