Let’s explain step by step 🐇

how to reverse part of a linked list from left = 2 to right = 4 using an example 1:

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Input:

The linked list: 1 -> 2 -> 3 -> 4 -> 5 We need to reverse the section between positions left = 2 and right = 4.

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Step 1: Initialization

Create a dummy node to make edge cases easier to handle. The list now looks like this:

dummy -> 1 -> 2 -> 3 -> 4 -> 5

Declare pointers:

• prev: Points to the node just before the left position. Initially, prev points to dummy.

• Move prev to the node just before left. After the loop, prev points to:

prev -> 1

Identify important nodes:

• start: The first node of the section to be reversed (left position), which is 2.

• then: The node immediately after start, which is 3.

Initial setup looks like this:

dummy -> 1 -> 2 -> 3 -> 4 -> 5
           prev   start   then

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Step 2: Reverse the Sublist

We loop right - left times (2 times in this case) to reverse the section between left and right.

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Iteration 1:

Update links:

• start.next = then.next: Node 2 (start) now points to 4 (skipping 3).

• then.next = prev.next: Node 3 (then) now points to 2.

• prev.next = then: Node 1 (prev) now points to 3.

The list now looks like this:

dummy -> 1 -> 3 -> 2 -> 4 -> 5
         prev   then   start

Move then to the next node after start (node 4):

then -> 4

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Iteration 2:

Update links:

• start.next = then.next: Node 2 (start) now points to 5 (skipping 4).

• then.next = prev.next: Node 4 (then) now points to 3.

• prev.next = then: Node 1 (prev) now points to 4.

The list now looks like this:

dummy -> 1 -> 4 -> 3 -> 2 -> 5
         prev         start

Move then to the next node after start (node 5):

theen -> 5

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Step 3: Return the Result

After completing the loop, the section from left to right has been reversed. Return the list starting from dummy.next:

1 -> 4 -> 3 -> 2 -> 5