LESSON 39 · INTERMEDIATE4 min read

Pointer Manipulation: Iterative Inversion and Reverse Nodes in K-Group

Master iterative and recursive in-place pointer manipulation: reversing singly linked lists, reversing in k-groups, and checking palindromes with O(1) space.

OpenJDK 21+ (LTS)Free to readModule 10 of 15
YOUR NEXT SMALL WINS

By the end, you’ll be able to…

  • Reverse a singly linked list in-place using a three-pointer sliding window without allocating nodes
  • Implement Reverse Nodes in K-Group by coordinating sublist boundaries and stitching outer pointers
  • Verify whether a linked list is a palindrome in O(n) time and O(1) auxiliary space

01Three-Pointer Sliding Window for In-Place Reversal

Reversing a singly linked list in-place is one of the most fundamental pointer manipulation skills in systems programming. The core challenge is that once a node's next pointer is overwritten to point backward, the reference to the remainder of the list is severed.

To prevent lost pointers, we maintain three pointers: prev (initially null), curr (initially head), and nextTemp. At each step, nextTemp saves curr.next. Then curr.next is redirected to prev. Finally, prev moves to curr, and curr advances to nextTemp.

This invariant ensures that no heap memory is allocated (achieving O(1) auxiliary space), and every node is visited exactly once (achieving O(n) time complexity).

example.js☕Java
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02Step-by-Step Pointer Trace Table

Tracing the three pointers over a 4-node list [1 -> 2 -> 3 -> 4 -> null] illustrates how links are systematically inverted without breaking continuity:

notes.txtPlain text
+------+------+----------+----------+-----------------------+-----------------------+| Step | prev | curr.val | nextTemp | Link Modified         | State After Step      |+------+------+----------+----------+-----------------------+-----------------------+| Init | null | 1        | -        | None                  | 1 -> 2 -> 3 -> 4      || 1    | 1    | 2        | 2        | 1.next = null         | null <- 1  2 -> 3 -> 4|| 2    | 2    | 3        | 3        | 2.next = 1            | null <- 1 <- 2  3 -> 4|| 3    | 3    | 4        | 4        | 3.next = 2            | null <- 1 <- 2 <- 3  4|| 4    | 4    | null     | null     | 4.next = 3            | null <- 1 <- 2 <- 3 <- 4+------+------+----------+----------+-----------------------+-----------------------+Final Result: prev points to 4 (the new head: 4 -> 3 -> 2 -> 1 -> null).

03Reverse Nodes in K-Group: Coordinating Sublist Stitching

Reverse Nodes in K-Group represents the pinnacle of linked list pointer problems. Given a linked list, reverse the nodes of the list k at a time, and return its modified list. If the number of nodes is not a multiple of k, left-out nodes at the end should remain in their original order.

We solve this cleanly using a dummy node and a groupPrev pointer. First, we scan ahead k nodes to verify that a full group exists. If fewer than k nodes remain, we terminate.

Otherwise, we isolate the k-group, reverse its internal pointers, and reconnect groupPrev.next to the group's new head, and the original group's head to the next group's tail. We then advance groupPrev to the end of the newly reversed group.

example.js☕Java
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04Palindrome Verification in O(1) Space

To check if a singly linked list is a palindrome without copying values into an ArrayList: 1. Use slow and fast pointers to find the middle node. 2. Reverse the second half of the list in-place.

3. Compare values element-by-element between the first half and the reversed second half. 4. (Optional but good practice) Restore the list by reversing the second half back before returning.

This algorithm achieves strict O(n) runtime and O(1) auxiliary space, adhering to production systems requirements.

05Production Architecture & Engineering Trade-Offs

Algorithmic efficiency in real-world software is heavily influenced by CPU hardware architecture, specifically the L1, L2, and L3 cache hierarchies and hardware branch predictors. While an algorithm may exhibit an optimal theoretical asymptotic complexity like O(n), poor spatial locality caused by pointer chasing across fragmented memory blocks can make it perform slower than a theoretically inferior algorithm operating over contiguous array memory.

When implementing two pointers and sliding window routines, developers must maintain strict boundary invariants to eliminate off-by-one errors and avoid redundant recomputations. Precomputing prefix sums or maintaining monotonic tracking queues transforms polynomial search operations into deterministic linear scans that execute with minimal memory allocations.

LET IT CLICK

The ideas worth keeping.

  • Three-pointer manipulation reverses links in O(n) time and O(1) space without data copying.
  • Reverse in K-Group coordinates sublist boundary pointers and stitches them via sentinel nodes.
  • Palindrome checking combines midpoint discovery with second-half reversal to achieve O(1) memory.
TURN UNDERSTANDING INTO SOMETHING YOU CAN DO

Make it yours.

Type and test your code right here in the interactive workspace. Run your solution, check the output, and compare with the worked guide when ready.

01

Exercise 1: Implement the middle finding logic using slow

Implement the middle-finding logic using slow and fast pointers where fast advances by 2 and slow advances by 1.

your_turn_1.java☕Java
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Output
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Explore a worked solution
one_possible_solution.javaJava
public class FindMiddle {    public static ListNode getMiddle(ListNode head) {        ListNode slow = head;        ListNode fast = head;        while (fast != null && fast.next != null) {            slow = slow.next;            fast = fast.next.next;        }        return slow;    }}

When fast reaches the end, slow is positioned exactly at the middle node.

02

Exercise 2: Explain why using a dummy node simplifies lin

Explain why using a dummy node simplifies linked list reversal problems involving head modifications.

your_turn_2.java☕Java
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Explore a worked solution
one_possible_solution.javaJava
// A dummy node (sentinel) placed before the head eliminates special-case branching// when the head node itself must be replaced, removed, or reversed.// By referencing dummy.next, the calling code can always retrieve the final head cleanly// without null checks or conditional head assignment logic.

Think about what happens when the first element of the list changes.

A SMALL QUESTION. A USEFUL NEXT STEP.

What stayed with you?

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A LITTLE ACTIVE RECALL0 / 3 answered
01What is the auxiliary space complexity of in-place linked list reversal?
02In Reverse Nodes in K-Group, what must happen if the remaining nodes at the end are fewer than k?
03How do you detect the middle of a linked list in a single pass?
⚡CODING CHALLENGE
🔒Required to Mark Lesson Complete

Exercise 1: Implement the middle finding logic using slow

Implement the middle-finding logic using slow and fast pointers where fast advances by 2 and slow advances by 1.

EXPECTED OUTPUT
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