LeetCode #444. Sequence Reconstruction

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探讨如何通过一系列子序列重构原始序列，并确保其唯一性。利用拓扑排序算法，检查子序列是否能唯一确定一个由1到n的排列，同时验证所有数字是否在指定范围内。

题目描述：

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.

class Solution {
public:
    bool sequenceReconstruction(vector<int>& org, vector<vector<int>>& seqs) {
        //根据所有的子序列建图，那么利用拓扑排序，每次有且只能有一个入度为零的点，否则就不能保证唯一性
        //另外还要判断所有子序列的数字都在1-n的范围内，同时包含1-n的所有数字
        int n=org.size();
        unordered_map<int,unordered_set<int>> graph;
        unordered_map<int,int> indegree;
        for(vector<int>& seq:seqs)
        {
            for(int i=0;i<seq.size();i++)
            {
                int u=seq[i];
                if(u<=0||u>n) return false;
                if(indegree.count(u)==0) indegree[u]=0;
                if(graph.count(u)==0) graph[u]={};
                if(i<seq.size()-1)
                {
                    int v=seq[i+1];   
                    if(v<=0||v>n) return false; // seqs的数字超出范围
                    if(graph[u].count(v)) continue; // 避免重边
                    graph[u].insert(v);
                    indegree[v]++;
                }
            }
        }
        
        queue<int> q;
        for(int i=1;i<=n;i++)
        {
            if(indegree.count(i)==0) return false; // seqs中没有i
            if(indegree[i]==0) q.push(i);
        }
        
        vector<int> result;
        while(!q.empty())
        {
            if(q.size()!=1) return false;
            int i=q.front();
            q.pop();
            result.push_back(i);
            for(int neighbor:graph[i])
            {
                indegree[neighbor]--;
                if(indegree[neighbor]==0) q.push(neighbor);
            }
            indegree.erase(i);
            graph.erase(i);
        }
        return result==org;
    }
};