How can the runner technique be used to arrange this linked list?

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In summary, the conversation was discussing the positioning of two pointers, p1 and p2, in a linked list. The person was unsure of how p2 would be at the midpoint when p1 reaches the end of the list. They provided a thought experiment with n representing the number of nodes in the list, and concluded that n must be even for this to work. It was then clarified that n can be even or odd, as long as it is half the length of the list.
  • #1
ainster31
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Homework Statement



POpQu.png


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The Attempt at a Solution



I don't understand how when p1 hits the end of the linked list, p2 will be at the midpoint.

This is how I am imagining it if n=3. Each step below represents an iteration.

1. a1 (p1, p2)->a2->a3->b1->b2->b3

2. a1->a2 (p2)->a3 (p1)->b1->b2->b3

3. a1->a2->a3 (p2)->b1->b2 (p1)->b3

4. Index out of bounds error because p2 now points to a node after b3.
 
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  • #2
It says you don't know the list length but you do know that its length is even.

Hence you can't have n=3 in your thought experiment.

So if n=10 then:

i=0 p1=0 p2=2
i=1 p1=1 p2=4
i=2 p1=2 p2=6
i=3 p1=3 p2=8
i=4 p1=4 p2=0 // p2 is reset to the beginning of the list and p1 is at the midpoint
 
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  • #3
Oh, so I guess they wanted n to be even. Even if n is odd, the linked list length is even since the length of the linkedlist is 2n.
 
  • #4
ainster31 said:
Oh, so I guess they wanted n to be even. Even if n is odd, the linked list length is even since the length of the linkedlist is 2n.

Only the linked list length is even, if n is half the list length then n can be even or odd.
 
  • #5


The runner technique can be used to arrange a linked list by using two pointers, p1 and p2, where p1 moves at a slower pace than p2. This approach can be used to find the midpoint of the linked list by having p1 move one node at a time and p2 move two nodes at a time. When p2 reaches the end of the linked list, p1 will be at the midpoint. This can be useful for sorting the linked list by splitting it into two smaller lists at the midpoint, sorting each list separately, and then merging them back together in sorted order. Alternatively, the runner technique can also be used to check for cycles in the linked list by having p1 and p2 move at different speeds and checking if they ever meet again.
 

1. How does the runner technique work for arranging a linked list?

The runner technique involves using two separate pointers, a "slow" pointer and a "fast" pointer, to traverse through the linked list. The slow pointer moves through the list one node at a time, while the fast pointer moves through the list two nodes at a time. This allows us to keep track of the current and previous nodes as we rearrange the list.

2. What are the benefits of using the runner technique for arranging a linked list?

The runner technique is a more efficient method for arranging a linked list compared to other techniques, such as using recursion or creating a new list. It also allows us to rearrange the list in-place without requiring extra memory space.

3. How do you determine when to stop using the runner technique for arranging the linked list?

The runner technique should stop when the fast pointer reaches the end of the list, indicating that the slow pointer has reached the middle of the list. At this point, we can start rearranging the nodes in the list.

4. Can the runner technique be used for any type of linked list?

Yes, the runner technique can be used for any type of linked list, including singly linked lists, doubly linked lists, and circular linked lists. It is a versatile technique that can be applied to various data structures.

5. How can the runner technique be implemented in code for arranging a linked list?

To implement the runner technique in code, we first declare two pointers, "slow" and "fast", and set them to the head of the linked list. We then use a while loop to move the pointers through the list until the fast pointer reaches the end. Once the loop is complete, we can rearrange the nodes using the current and previous nodes tracked by the pointers.

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