Swap Nodes in Pairs
A medium Linked List problem included in Striver A2Z. Below: the roles whose interviews prioritise this topic, and how to practise it.
- Topic
- Linked List
- Sheets
- 1
- Core for
- 3 roles
- Platform
- LeetCode
The problem
Given a linked list, swap every two adjacent nodes and return its head.
Example 1
- Input
- [1,2,3,4]
- Output
- [2,1,4,3]
Example 2
- Input
- []
- Output
- []
Example 3
- Input
- [1]
- Output
- [1]
Constraints
- 0 <= n <= 100
- -100 <= Node.val <= 100
How to think about it
Updated 2026-09-09A swap between two nodes affects three connections: the predecessor pointing into the pair, the link between the two nodes themselves, and the successor trailing after them. Anchoring with a dummy node ensures that reordering the very first pair uses the exact same pointer logic as every subsequent pair.
Approaches, worst first
Recursive pairwise swap
time O(n) · space O(n)
Swap the first two nodes directly, and recursively link `first.next` to the result of swapping the remaining tail. Clean and concise, but allocates call stack frames proportional to list length.
Iterative pointer rewiring with dummyWrite this one
time O(n) · space O(1)
Anchor with a dummy node before head and point `prev` to it. While `prev.next` and `prev.next.next` exist, label them `first` and `second`. Set `prev.next = second`, `first.next = second.next`, and `second.next = first`. Advance `prev` to `first` and repeat.
Where people lose marks · 3
- Swapping node values instead of actual pointer links bypasses the mechanical requirement of the problem and fails if node identity matters.
- Losing reference to `second.next` before assigning `second.next = first` leaves the remainder of the list permanently unreachable.
- Lists with 0 or 1 node must return head immediately without attempting pair dereferences.
The theory behind it
Linked List — the ground this problem stands on. All Linked List problems
What Linked List is
A linked list is a chain of separate cargo cars connected by coupling hooks, scattered anywhere across memory rather than sitting in a tidy contiguous row. Each car, called a node, holds a single piece of data and a pointer directing traffic to the address of the next car in line. Because nodes connect only by directional links, jumping straight to the tenth car is impossible without walking past the first nine.
When to reach for it
Choose a linked list when a problem requires frequent insertions and deletions at known positions without shifting whole blocks of surrounding memory. Problems mentioning pointer splicing, reversing subsequences in place, merging sorted streams, or detecting cycles in linear chains strongly point here. It is ideal when total capacity is unpredictable and memory allocation must happen one individual node at a time.
How the pattern works
Think in terms of pointer rewiring before dereferencing. Keep a dummy head node pointing to the start of the list so modifications to the initial item do not require separate edge logic. Always save references to neighboring nodes into temporary variables before cutting or redirecting forward links. When diagnosing loops or locating middle nodes, advance two references simultaneously at differing velocities so traversal completes without supplementary storage.
What each operation costs
| Operation | Time |
|---|---|
| insert or delete at the head | O(1) |
| insert or delete after a known node | O(1) |
| find an element by value or position | O(n) |
What usually goes wrong with Linked List
- Losing access to the remainder of the chain by overwriting a next reference before caching the downstream node address in a temporary variable.
- Attempting to read properties of a null node reference after walking one step beyond the tail or advancing a fast runner without checking its next step.
- Creating an accidental infinite cycle by pointing a trailing node back into earlier segments of the chain without severing old outgoing links.
Which roles need this problem
Linked List is a core topic for these 3 roles — if you're targeting one of them, this problem is early in your path, not optional.
Secondary for 5 more roles, including Full-Stack Developer, Android Developer, iOS Developer.
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