MHB Can we merge two unsorted lists in constant time?

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Merging two unsorted lists in constant time requires maintaining pointers to both the head and tail of each list. By ensuring that the tail of the first list points to the head of the second list, the merge operation can be completed in O(1) time. This approach avoids the need to traverse the lists, which would otherwise result in O(n) complexity. It is essential to keep track of these pointers correctly to facilitate efficient merging. Ultimately, the algorithm can return a pointer to the head of the merged list.
  • #31
I like Serena said:
That's it. (Mmm)

Great! Thank you! (Party)
 
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  • #32
Code: 
L1_tail.next=L2.head;
L1_tail=L2_tail;

If we would have an array $A$ and we would use a sentinel node, would it be like that? (Thinking)

Code: 
A[i].sentinel=A[j].list;
A[i].sentinel=A[j].sentinel
 

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