Reorder Linked List in DSA Go - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Reorder Linked List problem?

To rearrange a linked list so that nodes are ordered as first node, last node, second node, second last node, and so on.

intermediate

Which three main steps are used to reorder a linked list?

1. Find the middle of the list.
2. Reverse the second half.
3. Merge the two halves alternating nodes.

beginner

How do you find the middle of a linked list efficiently?

Use two pointers: a slow pointer moving one step at a time and a fast pointer moving two steps. When fast reaches the end, slow is at the middle.

intermediate

Why do we reverse the second half of the linked list in the reorder process?

Reversing the second half allows easy merging from the end towards the middle, creating the desired alternating order.

intermediate

What is the time complexity of the reorder linked list algorithm?

O(n), where n is the number of nodes, because each step (finding middle, reversing, merging) processes nodes linearly.

What does the fast pointer do when finding the middle of a linked list?

A Stays at the head

B Moves one step at a time

C Moves two steps at a time

D Moves three steps at a time

The fast pointer moves two steps at a time to reach the end faster, helping find the middle with the slow pointer.

After finding the middle, what is the next step in reordering the linked list?

A Reverse the first half

B Delete the middle node

C Merge the two halves immediately

D Reverse the second half

We reverse the second half to prepare for merging nodes from the end towards the middle.

What is the final step in the reorder linked list algorithm?

A Reverse the entire list

B Merge the two halves alternating nodes

C Find the middle node

D Sort the list

The final step merges the first half and reversed second half by alternating nodes to achieve the reorder.

What is the space complexity of the reorder linked list algorithm?

A O(1)

B O(log n)

C O(n)

D O(n^2)

The algorithm uses constant extra space by rearranging pointers without extra data structures.

If the linked list is 1 -> 2 -> 3 -> 4 -> 5, what is the reordered list?

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

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

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

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

The reordered list alternates first and last nodes: 1, then 5, then 2, then 4, then 3.

Explain the three main steps to reorder a linked list and why each step is important.

Think about how to split, reverse, and then combine the list.

Describe how the two-pointer technique helps in the reorder linked list problem.

Focus on how to find the middle node quickly.

Practice

1. What is the main goal of the Reorder Linked List problem?

easy

A. Rearrange the list so nodes alternate from start and end towards the center

B. Sort the linked list in ascending order

C. Reverse the entire linked list

D. Remove duplicate nodes from the linked list

Solution

1. Step 1: Understand the problem statement

   The problem asks to reorder the list so that nodes come alternately from the start and the end, like L0 -> Ln -> L1 -> Ln-1 -> L2 -> Ln-2 and so on.

2. Step 2: Differentiate from other operations

   This is not sorting or reversing the entire list, but a specific pattern of rearrangement.

3. Final Answer:

   Rearrange the list so nodes alternate from start and end towards the center -> Option A

4. Quick Check:

   Reorder = alternate start/end nodes [OK]

Hint: Think of weaving nodes from front and back alternately [OK]

Common Mistakes:

• Confusing reorder with sorting
• Trying to reverse the whole list only
• Ignoring the alternating pattern

2. Which of the following Go code snippets correctly declares a singly linked list node struct for the reorder linked list problem?

easy

A. class ListNode { int val; ListNode next; }

B. type ListNode struct { Val int; Next *ListNode }

C. struct ListNode { int Val; ListNode* Next; }

D. type ListNode struct { Value int; NextNode *ListNode }

Solution

1. Step 1: Identify Go struct syntax

   In Go, structs are declared with the keyword type, fields have names and types separated by spaces, and pointers use *.

2. Step 2: Match correct field names and types

   type ListNode struct { Val int; Next *ListNode } uses Val int and Next *ListNode, which is the standard for singly linked list nodes in Go.

3. Final Answer:

   type ListNode struct { Val int; Next *ListNode } -> Option B

4. Quick Check:

   Go struct with pointer field = type ListNode struct { Val int; Next *ListNode } [OK]

Hint: Remember Go struct fields: name type; pointer uses * [OK]

Common Mistakes:

• Using C/C++ or Java syntax in Go
• Missing pointer * for next node
• Incorrect field names or types

3. Given the linked list 1 -> 2 -> 3 -> 4 -> 5 -> null, what is the reordered list after applying the reorder linked list algorithm?

medium

A. 1 -> 5 -> 2 -> 4 -> 3 -> null

B. 5 -> 4 -> 3 -> 2 -> 1 -> null

C. 1 -> 2 -> 3 -> 4 -> 5 -> null

D. 1 -> 3 -> 5 -> 2 -> 4 -> null

Solution

1. Step 1: Identify the reorder pattern

   The reorder pattern takes the first node, then the last node, then second node, then second last node, and so on.

2. Step 2: Apply pattern to given list

   Start: 1, end: 5, then 2, then 4, then 3 in the middle. So the reordered list is 1 -> 5 -> 2 -> 4 -> 3 -> null.

3. Final Answer:

   1 -> 5 -> 2 -> 4 -> 3 -> null -> Option A

4. Quick Check:

   Alternate front and back nodes = 1 -> 5 -> 2 -> 4 -> 3 -> null [OK]

Hint: Pick nodes alternately from front and back [OK]

Common Mistakes:

• Reversing the entire list instead
• Keeping original order unchanged
• Mixing node order incorrectly

4. Consider this Go function snippet for reordering a linked list:

func reorderList(head *ListNode) {
    if head == nil || head.Next == nil { return }
    slow, fast := head, head
    for fast != nil && fast.Next != nil {
        slow = slow.Next
        fast = fast.Next.Next
    }
    second := slow.Next
    slow.Next = nil
    var prev *ListNode
    for second != nil {
        next := second.Next
        second.Next = prev
        prev = second
        second = next
    }
    first, second = head, prev
    for second != nil {
        tmp1 := first.Next
        tmp2 := second.Next
        first.Next = second
        second.Next = tmp1
        first = tmp1
        second = tmp2
    }
}

What is the bug in this code?

medium

A. The function does not handle empty or single-node lists

B. The middle node is not correctly found; slow pointer should start at head.Next

C. The loop to reverse the second half does not update the prev pointer correctly

D. The reversed second half is not properly linked back to the first half

Solution

1. Step 1: Analyze the reversal and merging steps

   The code correctly finds the middle, reverses the second half, and then merges nodes alternately.

2. Step 2: Identify the bug in linking halves

   After merging, the last node of the reordered list may still point to old nodes if not handled carefully. The code misses ensuring the last node's Next is nil, causing potential cycle or dangling links.

3. Final Answer:

   The reversed second half is not properly linked back to the first half -> Option D

4. Quick Check:

   Linking halves correctly is key [OK]

Hint: Check if last node's Next is set to nil after merge [OK]

Common Mistakes:

• Not breaking the list at middle properly
• Forgetting to set slow.Next = nil
• Incorrect pointer updates in merge loop

5. You have a linked list with nodes containing values 1 -> 2 -> 3 -> null. After reordering, what will be the list if you apply the reorder linked list algorithm twice in a row?

hard

A. 1 -> 3 -> 2 -> null (same as after first reorder)

B. 3 -> 2 -> 1 -> null

C. 1 -> 2 -> 3 -> null (original list)

D. 2 -> 3 -> 1 -> null

Solution

1. Step 1: Apply reorder once

   First reorder transforms 1->2->3 into 1->3->2.

2. Step 2: Apply reorder again on the new list

   Reordering 1->3->2 again returns the original order 1->2->3 because the reorder pattern is symmetric and reverses itself on second application.

3. Final Answer:

   1 -> 2 -> 3 -> null (original list) -> Option C

4. Quick Check:

   Reorder twice returns original list [OK]

Hint: Reordering twice restores original order [OK]

Common Mistakes:

• Assuming reorder twice keeps the first reorder result
• Not recognizing reorder symmetry
• Confusing reorder with reverse