444.Sequence-Reconstruction

444. Sequence Reconstruction

题目地址

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
Constraints:
1 <= n <= 10^4
org is a permutation of {1,2,...,n}.
seqs[i][j] fits in a 32-bit signed integer.

代码

Approach #1 BFS Topological Sort

This question can be simplified to three conditions:
  1. 1.
    TopSort order exists
  2. 3.
    the only top sort order constructed should be equal to the org.
index == org.length (check condition 3) && index == map.size() (check all the vertex in the graph has been visited, so the top sort order exists, check condition 1)
How to check only one order? queue.size() should always be one, then only one element at a time has indegree to be 0, so you only have one choice (check condition 2)
class Solution {
public boolean sequenceReconstruction(int[] org, List<List<Integer>> 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;
whhile (!queue.isEmtpy()) {
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(); // 无环,所有顶点被遍历到
}
}