Linked List In-Place Manipulation

Fundamentals

Reversing, reordering, and partitioning pointer chains with O(1) space

70 problems·13 Easy·50 Medium·7 Hard
Pattern Study Guide & Cheat Sheet

Rearrange nodes and pointers in-place by maintaining dummy heads, tracking previous/current/next pointers, and preserving detached segments.

Core Invariant: Pointer Surgery: Always save `temp = curr.next` before overwriting `curr.next`. Use a `dummy` node (`dummy.next = head`) to eliminate edge-case conditionals when mutating the head.

Recognize it (Keywords & Signals)
  • Reverse linked list / reverse nodes in K-groups
  • Merge two or K sorted linked lists
  • Remove N-th node from end of list
  • Reorder list (interleave first half with reversed second half)
  • Partition list around value X
When NOT to use

Random indexed access is required or node count is static and O(N) space is acceptable (convert to list).

How to solve (Step-by-step)
  1. 1.Create `dummy = ListNode(0, head)` and set `prev = None`, `curr = head`.
  2. 2.Before modifying any link: save `next_node = curr.next`.
  3. 3.Reverse or rewire pointer: `curr.next = prev`.
  4. 4.Advance pointers: `prev = curr; curr = next_node`.
  5. 5.Return `dummy.next` or `prev` depending on whether head was replaced.
Watch for (Interview Traps)
  • Overwriting `curr.next` before caching the old next reference
  • Creating circular loops by forgetting to terminate `tail.next = None`
  • Failing on empty list (`head = None`) or single-node lists
Iterative In-Place Linked List Reversal
def reverse_linked_list(head: ListNode | None) -> ListNode | None:
    prev = None
    curr = head
    
    while curr:
        next_temp = curr.next  # 1. Save next reference before breaking link
        curr.next = prev       # 2. Reverse direction of pointer
        prev = curr            # 3. Advance prev forward
        curr = next_temp       # 4. Advance curr forward
        
    return prev  # New head of reversed list
Cost
O(n) single pass · O(1) in-place pointer updates (Zero node allocations; all modifications are in-place.)
Canonical problems
#206 Reverse Linked List: Foundational 3-pointer reversal
#21 Merge Two Sorted Lists: Dummy head + compare node values to splice
#143 Reorder List: Find middle + reverse second half + interleave
#25 Reverse Nodes in k-Group: Count k nodes, reverse segment, reconnect recursively
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