Creating a Singly Linked List from Scratch in DSA Go - Code It Step by Step

Creating a Singly Linked List from Scratch

📖 Scenario: Imagine you want to keep track of your daily tasks in order. A singly linked list is like a chain where each task points to the next one. You will build this chain step-by-step in Go.

🎯 Goal: Build a simple singly linked list in Go that holds three tasks and prints them in order.

📋 What You'll Learn

Define a Node struct with data and next fields

Create three nodes with exact task names: "Task1", "Task2", "Task3"

Link the nodes to form a chain

Print the tasks in order separated by -> and ending with nil

💡 Why This Matters

🌍 Real World

Singly linked lists are used in many applications like task scheduling, undo functionality, and managing playlists where order matters.

💼 Career

Understanding linked lists is fundamental for software developers, especially for roles involving system design, memory management, and algorithm optimization.

1

Define the Node struct and create the first node

Define a struct called Node with two fields: data of type string and next of type *Node. Then create a variable called head of type *Node and assign it a new Node with data set to "Task1" and next set to nil.

DSA Go

package main

type Node struct {
    data string
    next *Node
}

func main() {
    // Create the head node with data "Task1" and next nil
    // Your code here
}

Hint

Remember, next points to the next node or nil if none.

2

Create the second and third nodes and link them

Create two more variables called second and third of type *Node. Assign second a new Node with data set to "Task2" and next set to nil. Assign third a new Node with data set to "Task3" and next set to nil. Then link the nodes by setting head.next to second and second.next to third.

DSA Go

package main

type Node struct {
    data string
    next *Node
}

func main() {
    head := &Node{data: "Task1", next: nil}
    // Create second and third nodes with data "Task2" and "Task3"
    // Link head.next to second and second.next to third
    // Your code here
}

Hint

Each node's next points to the following node.

3

Traverse the linked list and collect task names

Create a variable called current and set it to head. Create a slice of strings called tasks to store the task names. Use a for loop that continues while current is not nil. Inside the loop, append current.data to tasks and then set current to current.next.

DSA Go

package main

type Node struct {
    data string
    next *Node
}

func main() {
    head := &Node{data: "Task1", next: nil}
    second := &Node{data: "Task2", next: nil}
    third := &Node{data: "Task3", next: nil}

    head.next = second
    second.next = third

    // Traverse the list starting from head
    // Append each node's data to tasks slice
    // Your code here
}

Hint

Use for current != nil to loop through the nodes.

4

Print the linked list tasks in order

Use a for loop with index i to iterate over the tasks slice. For each task except the last, print the task followed by " -> " without a newline. For the last task, print the task followed by " -> nil" with a newline.

DSA Go

package main

type Node struct {
    data string
    next *Node
}

func main() {
    head := &Node{data: "Task1", next: nil}
    second := &Node{data: "Task2", next: nil}
    third := &Node{data: "Task3", next: nil}

    head.next = second
    second.next = third

    current := head
    tasks := []string{}
    for current != nil {
        tasks = append(tasks, current.data)
        current = current.next
    }

    // Print tasks in order separated by " -> " and ending with " -> nil"
    // Your code here
}

Hint

Use fmt.Print for tasks except the last, and fmt.Println for the last task.

Practice

1. What is the main purpose of a singly linked list in data structures?

easy

A. To store elements in a hash table for fast lookup

B. To store elements in a tree-like structure with multiple branches

C. To store elements in a fixed-size array for quick access

D. To store elements in a linear order where each element points to the next

Solution

1. Step 1: Understand singly linked list structure

   A singly linked list stores elements in nodes where each node points to the next node, forming a chain.

2. Step 2: Compare with other data structures

   Unlike trees or arrays, singly linked lists have a linear sequence with pointers only to the next element.

3. Final Answer:

   To store elements in a linear order where each element points to the next -> Option D

4. Quick Check:

   Singly linked list = linear chain with next pointers [OK]

Hint: Remember: singly linked list nodes point only forward [OK]

Common Mistakes:

• Confusing singly linked list with trees or arrays
• Thinking nodes have pointers to previous nodes
• Assuming fixed size storage like arrays

2. Which of the following is the correct way to define a node struct for a singly linked list in Go?

easy

A. type Node struct { value int; next *Node }

B. type Node struct { value int; prev *Node }

C. type Node struct { value int; next Node }

D. type Node struct { value int; next interface{} }

Solution

1. Step 1: Identify correct pointer usage

   In Go, the next node must be a pointer to Node, so *Node is correct.

2. Step 2: Check struct fields

   Field 'value' stores data, 'next' points to the next node. 'prev' is for doubly linked lists, and 'interface{}' is too generic.

3. Final Answer:

   type Node struct { value int; next *Node } -> Option A

4. Quick Check:

   Node next pointer = *Node [OK]

Hint: Use *Node for next pointer in Go struct [OK]

Common Mistakes:

• Using value type instead of pointer for next
• Confusing singly linked list with doubly linked list (prev pointer)
• Using interface{} instead of *Node for next

3. What will be the output of the following Go code that creates a singly linked list with nodes 1 -> 2 -> 3?

package main
import "fmt"
type Node struct {
  value int
  next  *Node
}
func main() {
  head := &Node{value: 1}
  second := &Node{value: 2}
  third := &Node{value: 3}
  head.next = second
  second.next = third
  current := head
  for current != nil {
    fmt.Print(current.value)
    if current.next != nil {
      fmt.Print(" -> ")
    }
    current = current.next
  }
  fmt.Println(" -> null")
}

medium

A. 1 -> 2 -> null

B. 3 -> 2 -> 1 -> null

C. 1 -> 2 -> 3 -> null

D. Compilation error

Solution

1. Step 1: Trace node creation and linking

   Nodes with values 1, 2, 3 are created and linked in order: head.next = second, second.next = third.

2. Step 2: Trace printing loop

   Loop prints current.value, then ' -> ' if next exists, moving from head to third, then prints ' -> null'.

3. Final Answer:

   1 -> 2 -> 3 -> null -> Option C

4. Quick Check:

   Print nodes in order 1,2,3 with arrows [OK]

Hint: Follow next pointers from head to end [OK]

Common Mistakes:

• Printing nodes in reverse order
• Missing last node in output
• Confusing pointer assignments causing nil errors

4. Identify the error in this Go code snippet that tries to append a new node with value 4 to the end of a singly linked list:

func appendNode(head *Node, val int) {
  newNode := &Node{value: val}
  current := head
  for current.next != nil {
    current = current.next
  }
  current = newNode
}

medium

A. The loop condition should be current != nil instead of current.next != nil

B. The new node is not linked to the list; current pointer is reassigned incorrectly

C. The new node should be created with next pointing to head

D. The function should return the new head pointer

Solution

1. Step 1: Analyze the loop and pointer update

   The loop correctly finds the last node where current.next is nil.

2. Step 2: Check how new node is linked

   Assigning current = newNode only changes the local variable, not the last node's next pointer. It should be current.next = newNode.

3. Final Answer:

   The new node is not linked to the list; current pointer is reassigned incorrectly -> Option B

4. Quick Check:

   Link new node by current.next = newNode [OK]

Hint: Assign new node to current.next, not current [OK]

Common Mistakes:

• Reassigning current instead of current.next
• Wrong loop condition causing nil pointer dereference
• Not returning updated head when needed

5. You want to create a function in Go that reverses a singly linked list. Which approach correctly reverses the list by changing the next pointers?

func reverseList(head *Node) *Node {
  var prev *Node = nil
  current := head
  for current != nil {
    nextTemp := current.next
    current.next = prev
    prev = current
    current = nextTemp
  }
  return prev
}

What will be the output if the original list is 1 -> 2 -> 3 -> null and you print the reversed list?

hard

A. 3 -> 2 -> 1 -> null

B. 1 -> 2 -> 3 -> null

C. null

D. Runtime error due to nil pointer

Solution

1. Step 1: Understand reversal logic

   The function iterates through the list, reversing each node's next pointer to point to the previous node.

2. Step 2: Trace reversal on list 1 -> 2 -> 3

   After reversal, the list becomes 3 -> 2 -> 1 -> null, and prev points to new head.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option A

4. Quick Check:

   Reversed list head = prev = 3 [OK]

Hint: Reverse pointers by tracking prev, current, next [OK]

Common Mistakes:

• Not updating current.next correctly
• Losing track of next node causing infinite loop
• Returning old head instead of new head