Reverse a linked list using Golang

Reverse a linked list using Golang

As we start coding and slowly learning about Data Structures(DS), we come across a very famous linear data structure which is known as a linked list. Linked lists and their related questions are quite popular in interviewers who love problem-solving.

What is a Linked List?

A linked list is a common linear data structure. its elements are also known as Node are not stored at a contiguous location. The nodes are linked using pointers. Linked lists are popular as their size is not fixed like an array. Linked list's each node contains a value and a pointer to the next node. The head pointer points to the first node, and the last node of the list points to null. When the linked list is empty, the head pointer points to null.

Types of Linked Lists:

  1. Singly Linked List (Uni-directional)
  2. Doubly Linked List (Bi-directional)
  3. Circular Linked List

I'll write about linked list types in a different post as this post about reversing a linked list.


1) Dynamically increase in size

2) Easy to insert/delete


1) Accessing a random node is not allowed.

2) Additional memory space needs for each node in a linked list.

3) Not cache-friendly.

Linked List

Node for a linked list

type Node struct {
    prev *Node
    next *Node
    key interface{}

type LinkedList struct {
    head *Node
    tail *Node

Push method for a Linked List

func (ll *LinkedList) Push(key interface{}) {
    list := &Node{
        next: ll.head,
        key: key,
    if ll.head != nil {
        ll.head.prev = list
    ll.head = list

    l := ll.head
    for != nil {
        l =
    ll.tail = l

Display a Linked list

func (ll *LinkedList) Display() {
    list := ll.head
    for list != nil {
        fmt.Printf("%+v ->", list.key)
        list =

// normal display function
func Display(list *Node) {
    for list != nil {
        fmt.Printf("%v ->", list.key)
        list =

Reverse a Linked list

func (ll *LinkedList) Reverse() {
    currentNode := ll.head
    var next *Node
    var previousNode *Node
    ll.tail = ll.head

    for currentNode != nil {
        next, =, previousNode
        previousNode, currentNode = currentNode, next
    ll.head = previousNode

If you like, you can read the same article on our official blog

Main function

func main() {
    link := LinkedList{}

    fmt.Printf("Head: %v\n", link.head.key)
    fmt.Printf("Tail: %v\n", link.tail.key)
    fmt.Printf("head: %v\n", link.head.key)
    fmt.Printf("tail: %v\n", link.tail.key)

// output
Head: 3
Tail: 9
3 -> 1 -> 8 -> 15 -> 12 -> 9 -> 
9 -> 12 -> 15 -> 8 -> 1 -> 3 -> 
head: 9
tail: 3