Reverse a Singly Linked List Iterative in DSA Java - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of reversing a linked list with 3 nodes

What is the printed state of the linked list after reversing it iteratively?

DSA Java

class Node {
    int val;
    Node next;
    Node(int val) { this.val = val; this.next = null; }
}

public class Main {
    public static Node reverseList(Node head) {
        Node prev = null;
        Node curr = head;
        while (curr != null) {
            Node nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }

    public static void printList(Node head) {
        Node curr = head;
        while (curr != null) {
            System.out.print(curr.val + " -> ");
            curr = curr.next;
        }
        System.out.print("null");
    }

    public static void main(String[] args) {
        Node head = new Node(1);
        head.next = new Node(2);
        head.next.next = new Node(3);
        Node reversed = reverseList(head);
        printList(reversed);
    }
}

A 3 -> 2 -> 1 -> null

B 1 -> 2 -> 3 -> null

C null

D 3 -> 1 -> 2 -> null

💡 Hint

Trace the pointers prev, curr, and nextTemp as the list reverses.

The iterative reversal changes the direction of the next pointers one by one. Starting from 1 -> 2 -> 3 -> null, after reversal it becomes 3 -> 2 -> 1 -> null.

❓ Predict Output

intermediate

Output when reversing an empty linked list

What is the output when the linked list is empty (head is null) and reversed iteratively?

DSA Java

class Node {
    int val;
    Node next;
    Node(int val) { this.val = val; this.next = null; }
}

public class Main {
    public static Node reverseList(Node head) {
        Node prev = null;
        Node curr = head;
        while (curr != null) {
            Node nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }

    public static void printList(Node head) {
        Node curr = head;
        while (curr != null) {
            System.out.print(curr.val + " -> ");
            curr = curr.next;
        }
        System.out.print("null");
    }

    public static void main(String[] args) {
        Node head = null;
        Node reversed = reverseList(head);
        printList(reversed);
    }
}

A null

B 0 -> null

C Exception thrown

D null -> null

💡 Hint

Consider what happens when the input list is empty.

If the input head is null, the reverse function returns null immediately. Printing null means no nodes are printed, just 'null'.

🔧 Debug

advanced

Identify the error in this iterative reverse code

What error will this code produce when reversing a linked list iteratively?

DSA Java

public static Node reverseList(Node head) {
    Node prev = null;
    Node curr = head;
    while (curr.next != null) {
        Node nextTemp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = nextTemp;
    }
    return prev;
}

A NullPointerException

B No error, works correctly

C Returns incomplete reversed list

D Infinite loop

💡 Hint

Check the loop condition and what happens to the last node.

The loop condition 'curr.next != null' stops before processing the last node. So the last node is never reversed or linked properly, resulting in an incomplete reversed list.

❓ Predict Output

advanced

Output after reversing a 5-node linked list

What is the printed linked list after reversing the list 10 -> 20 -> 30 -> 40 -> 50 iteratively?

DSA Java

class Node {
    int val;
    Node next;
    Node(int val) { this.val = val; this.next = null; }
}

public class Main {
    public static Node reverseList(Node head) {
        Node prev = null;
        Node curr = head;
        while (curr != null) {
            Node nextTemp = curr.next;
            curr.next = prev;
            prev = curr;
            curr = nextTemp;
        }
        return prev;
    }

    public static void printList(Node head) {
        Node curr = head;
        while (curr != null) {
            System.out.print(curr.val + " -> ");
            curr = curr.next;
        }
        System.out.print("null");
    }

    public static void main(String[] args) {
        Node head = new Node(10);
        head.next = new Node(20);
        head.next.next = new Node(30);
        head.next.next.next = new Node(40);
        head.next.next.next.next = new Node(50);
        Node reversed = reverseList(head);
        printList(reversed);
    }
}

A 10 -> 20 -> 30 -> 40 -> 50 -> null

B null

C 50 -> 30 -> 40 -> 20 -> 10 -> null

D 50 -> 40 -> 30 -> 20 -> 10 -> null

💡 Hint

Follow the reversal of pointers step-by-step for all nodes.

The iterative reversal reverses all next pointers, so the list 10 -> 20 -> 30 -> 40 -> 50 becomes 50 -> 40 -> 30 -> 20 -> 10 -> null.

🧠 Conceptual

expert

Why does iterative reversal use three pointers?

In iterative reversal of a singly linked list, why do we need to keep track of three pointers: prev, curr, and nextTemp?

A To create a new list instead of modifying the original

B To avoid losing access to the rest of the list when reversing links

C To speed up the reversal by skipping nodes

D To sort the list while reversing

💡 Hint

Think about what happens if you change a node's next pointer without saving the next node.

When reversing, changing curr.next breaks the link to the rest of the list. nextTemp saves the next node before changing the link, so we don't lose access. prev keeps track of the reversed part.

Practice

1. What is the main purpose of reversing a singly linked list iteratively?

easy

A. To sort the list in ascending order

B. To delete all nodes from the list

C. To change the direction of all links so the last node becomes the first

D. To add a new node at the end of the list

Solution

1. Step 1: Understand the linked list structure

   A singly linked list has nodes connected in one direction from head to tail.

2. Step 2: Purpose of reversing

   Reversing changes the links so the last node points to the previous one, making it the new head.

3. Final Answer:

   To change the direction of all links so the last node becomes the first -> Option C

4. Quick Check:

   Reversing = changing link directions [OK]

Hint: Reversing flips the list direction, last becomes first [OK]

Common Mistakes:

• Confusing reversing with sorting
• Thinking reversing deletes nodes
• Assuming reversing adds nodes

2. Which of the following is the correct way to initialize the previous pointer before starting the iterative reversal?

easy

A. Node prev = head;

B. Node prev = head.next;

C. Node prev = new Node();

D. Node prev = null;

Solution

1. Step 1: Understand the role of previous pointer

   Previous pointer starts as null because the new tail's next should be null after reversal.

2. Step 2: Initialization

   Setting prev to null ensures the first node processed points to null, ending the reversed list properly.

3. Final Answer:

   Node prev = null; -> Option D

4. Quick Check:

   Prev starts null to mark new tail [OK]

Hint: Prev pointer starts null to mark list end [OK]

Common Mistakes:

• Initializing prev to head causes cycle
• Creating new node wastes memory
• Starting prev at head.next skips first node

3. What will be the output linked list after reversing the list 1 -> 2 -> 3 -> null iteratively?

medium

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

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

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

D. 2 -> 3 -> 1 -> null

Solution

1. Step 1: Trace the reversal process

   Initially: 1 -> 2 -> 3 -> null. After reversal, links point backward: 3 -> 2 -> 1 -> null.

2. Step 2: Confirm final head and tail

   Node 3 becomes new head, node 1 becomes tail pointing to null.

3. Final Answer:

   3 -> 2 -> 1 -> null -> Option B

4. Quick Check:

   Reversed list = 3 -> 2 -> 1 [OK]

Hint: Last node becomes first after reversal [OK]

Common Mistakes:

• Assuming list stays same after reversal
• Confusing null placement
• Mixing node order

4. Identify the error in this iterative reversal code snippet:

Node prev = head;
Node current = head.next;
while (current != null) {
  Node nextNode = current.next;
  current.next = prev;
  prev = current;
  current = nextNode;
}
head.next = null;
head = prev;

medium

A. prev should be initialized to null, not head

B. current should start at head, not head.next

C. head.next should not be set to null

D. Loop condition should be current.next != null

Solution

1. Step 1: Check initialization of prev pointer

   Prev must start as null to mark new tail's next as null after reversal.

2. Step 2: Effect of wrong initialization

   Starting prev at head causes a cycle, breaking reversal logic.

3. Final Answer:

   prev should be initialized to null, not head -> Option A

4. Quick Check:

   Prev = null to avoid cycles [OK]

Hint: Prev must start null to avoid cycles [OK]

Common Mistakes:

• Starting prev at head causes infinite loop
• Wrong loop condition misses last node
• Forgetting to set old head next to null

5. Given a singly linked list with nodes containing values 10 -> 20 -> 30 -> 40 -> null, which iterative code snippet correctly reverses it and prints the reversed list?

hard

A. Node prev = null;
Node current = head;
while (current != null) {
Node next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
// Print list starting from head

B. Node prev = head;
Node current = head.next;
while (current != null) {
Node next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
// Print list starting from head

C. Node current = head;
while (current != null) {
Node next = current.next;
current.next = null;
current = next;
}
// Print list starting from head

D. Node prev = null;
Node current = head;
while (current.next != null) {
Node next = current.next;
current.next = prev;
prev = current;
current = next;
}
head = prev;
// Print list starting from head

Solution

1. Step 1: Verify initialization of pointers

   Prev must start as null, current at head to reverse all nodes correctly.

2. Step 2: Check loop condition and pointer updates

   Loop runs while current != null to process all nodes; inside loop, links reversed properly.

3. Step 3: Confirm final head assignment and printing

   After loop, head is set to prev (new head). Printing from head shows reversed list.

4. Final Answer:

   Option A code correctly reverses and prints the list -> Option A

5. Quick Check:

   Prev=null, loop full list, head=prev [OK]

Hint: Prev null, loop full list, head = prev after reversal [OK]

Common Mistakes:

• Starting prev at head causes cycle
• Loop condition misses last node
• Setting current.next = null inside loop breaks links