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Swapping Nodes in a Linked List

Medium Linked List Dummy Node Pattern Original on LeetCode

Swapping Nodes in a Linked List is a medium linked list problem solved with the dummy node pattern pattern. The best approach, optimal (one pass, front/back pointers), runs in O(n) time and O(1) space. Below are 2 approaches in Java, from two passes (count first) up.

Problem

Swap the values of the k-th node from the beginning and the k-th node from the end of a linked list (1-indexed), and return the head.

Examples

Example 1

Input
head = [7, 9, 6, 6, 8, 7, 3, 0, 9, 5], k = 5
Output
[7, 9, 6, 6, 7, 8, 3, 0, 9, 5]
Why
The 5th from the front (8) and the 5th from the end (7) swap values.

Example 2

Input
head = [1, 2, 3], k = 2
Output
[1, 2, 3]
Why
The middle node is both, so nothing changes.

Constraints

  • 1 <= k <= n <= 10^5
  • Swapping the values is allowed.

Animated walkthrough

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Solutions

Try it yourself first. Then compare: each approach lists its idea, the steps, its time and space, and the Java code.

ApproachTimeSpace
Two passes (count first)O(n)O(1)
Optimal (one pass, front/back pointers)O(n)O(1)

1Two passes (count first)

TimeO(n)
SpaceO(1)

Count n, then walk to node k and node n - k + 1 and swap their values.

  1. Count n.
  2. Find the k-th and (n - k + 1)-th nodes; swap values.
Java
class Solution {
    public ListNode swapNodes(ListNode head, int k) {
        int n = 0;
        for (ListNode p = head; p != null; p = p.next) n++;
        ListNode a = head, b = head;
        for (int i = 1; i < k; i++) a = a.next;
        for (int i = 1; i < n - k + 1; i++) b = b.next;
        int t = a.val; a.val = b.val; b.val = t;
        return head;
    }
}

2Optimal (one pass, front/back pointers)

TimeO(n)
SpaceO(1)

Walk first to the k-th node. Start second at the head and move both until first reaches the last node; second is then k-th from the end. Swap the values.

  1. first = k-th node; keep a reference to it.
  2. fast = first, second = head; while fast.next != null, move both.
  3. Swap values of the saved k-th node and second.
Java
class Solution {
    public ListNode swapNodes(ListNode head, int k) {
        ListNode first = head;
        for (int i = 1; i < k; i++) first = first.next;
        ListNode fast = first, second = head;
        while (fast.next != null) { fast = fast.next; second = second.next; }
        int t = first.val; first.val = second.val; second.val = t;
        return head;
    }
}

Edge cases to test

  • k points at the same node from both ends
  • k = 1 (swap head and tail)

Hints

Hint 1

Once a pointer is k steps ahead, move a second pointer from the head in lockstep until the first reaches the end.

FAQ

What is the best time complexity for Swapping Nodes in a Linked List?

Optimal (one pass, front/back pointers) runs in O(n) time and O(1) extra space.

Which pattern does Swapping Nodes in a Linked List use?

It is a linked list problem that uses the dummy node pattern pattern. Other problems with the same pattern: Remove Linked List Elements, Delete Nodes From Linked List Present in Array, Merge Two Sorted Lists.

Is there a brute force solution for Swapping Nodes in a Linked List?

Yes. Two passes (count first) takes O(n) time and O(1) space. Count n, then walk to node k and node n - k + 1 and swap their values.

Which edge cases should I test for Swapping Nodes in a Linked List?

k points at the same node from both ends; k = 1 (swap head and tail).