444. Sequence Reconstruction - jiejackyzhang/leetcode-note GitHub Wiki

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

##题目大意： 判断原始序列org能否由一组序列seqs重建。org序列是整数1到n的一个排列，其中1 ≤ n ≤ 10^4。重建意味着使用seqs中的序列构建一个最短公共超序列（也就是说，一个最短的序列满足seqs中的所有序列均为其子序列）。判断seqs中的序列能否唯一确定org序列。

##解题思路： 拓扑排序（Topological Sort）

将seqs中的各序列seq按照其中元素出现的先后顺序建立有向图g。

例如seqs中的某序列seq = [1, 2, 3]，对应有向图，顶点为1, 2, 3；边为(1, 2), (2, 3)。

利用Map<Integer, Integer> indegree记录各顶点的入度（indegree），Map<Integer, Set<Integer>> map记录各顶点的后继（边）。

然后对图g执行拓扑排序，将得到的排序结果与原始序列org作比对即可。

public class Solution {
    public boolean sequenceReconstruction(int[] org, int[][] seqs) {
        Map<Integer, Set<Integer>> map = new HashMap<>();
        Map<Integer, Integer> indegree = new HashMap<>();
        
        for(int[] seq: seqs) {
            if(seq.length == 1) {
                if(!map.containsKey(seq[0])) {
                    map.put(seq[0], new HashSet<>());
                    indegree.put(seq[0], 0);
                }
            } else {
                for(int i = 0; i < seq.length - 1; i++) {
                    if(!map.containsKey(seq[i])) {
                        map.put(seq[i], new HashSet<>());
                        indegree.put(seq[i], 0);
                    }

                    if(!map.containsKey(seq[i + 1])) {
                        map.put(seq[i + 1], new HashSet<>());
                        indegree.put(seq[i + 1], 0);
                    }

                    if(map.get(seq[i]).add(seq[i + 1])) {
                        indegree.put(seq[i + 1], indegree.get(seq[i + 1]) + 1);
                    }
                }
            }
        }

        Queue<Integer> queue = new LinkedList<>();
        for(Map.Entry<Integer, Integer> entry: indegree.entrySet()) {
            if(entry.getValue() == 0) queue.offer(entry.getKey());
        }

        int index = 0;
        while(!queue.isEmpty()) {
            int size = queue.size();
            if(size > 1) return false;
            int curr = queue.poll();
            if(index == org.length || curr != org[index++]) return false;
            for(int next: map.get(curr)) {
                indegree.put(next, indegree.get(next) - 1);
                if(indegree.get(next) == 0) queue.offer(next);
            }
        }
        return index == org.length && index == map.size();
    }
}
⚠️ **GitHub.com Fallback** ⚠️