Why Merge Two Sorted Linked Lists in DSA Rust?

The Big Idea

What if you could combine two sorted lists perfectly without lifting a finger?

The Scenario

Imagine you have two lists of names sorted alphabetically on paper. You want to combine them into one sorted list without messing up the order.

Doing this by hand means comparing each name one by one, which is slow and tiring.

The Problem

Manually comparing each item from two lists is slow and easy to make mistakes.

You might lose track of which name comes next or accidentally skip some names.

This gets worse as the lists get longer.

The Solution

Merging two sorted linked lists automatically compares and combines them step-by-step.

This method keeps the order correct without extra effort or mistakes.

It saves time and makes the process smooth and reliable.

Before vs After

✗ Before

let mut merged = Vec::new();
for name in list1 {
    merged.push(name);
}
for name in list2 {
    merged.push(name);
}
merged.sort();

✓ After

fn merge_lists(list1: Option<Box<Node>>, list2: Option<Box<Node>>) -> Option<Box<Node>> {
    // merge logic here
}

What It Enables

This lets you combine sorted data efficiently, enabling faster searching and processing in many programs.

Real Life Example

Combining two sorted contact lists on your phone into one without losing the alphabetical order.

Key Takeaways

Merging saves time by automating the combination of sorted lists.

It prevents errors common in manual merging.

Useful in many applications like databases and search engines.

Practice

1. What is the main goal when merging two sorted linked lists?

easy

A. Reverse both linked lists

B. Remove duplicates from both lists

C. Create a new sorted linked list containing all elements from both lists

D. Sort each list separately without merging

Solution

1. Step 1: Understand the problem

   Merging two sorted linked lists means combining them into one list that remains sorted.

2. Step 2: Identify the goal

   The goal is to create a new sorted linked list that contains all elements from both input lists.

3. Final Answer:

   Create a new sorted linked list containing all elements from both lists -> Option C

4. Quick Check:

   Merging = Combined sorted list [OK]

Hint: Merging means combining sorted lists into one sorted list [OK]

Common Mistakes:

• Thinking merging means reversing lists
• Trying to sort lists separately without merging
• Confusing merging with removing duplicates

2. Which Rust code snippet correctly declares a singly linked list node struct for merging?

easy

A. struct ListNode { val: i32, next: ListNode }

B. struct ListNode { val: i32, next: Option }

C. struct ListNode { val: i32, next: &ListNode }

D. struct ListNode { val: i32, next: Option<ListNode> }

Solution

1. Step 1: Understand Rust linked list node

   Rust needs heap allocation for recursive types, so next must be Option.

2. Step 2: Check each option

   struct ListNode { val: i32, next: Option } uses Option, which is correct. Others use references or direct types causing errors.

3. Final Answer:

   struct ListNode { val: i32, next: Option } -> Option B

4. Quick Check:

   Use Option for linked list nodes [OK]

Hint: Use Option for recursive linked list nodes in Rust [OK]

Common Mistakes:

• Using direct ListNode type for next causes infinite size error
• Using references without lifetimes causes borrow errors
• Using Option<ListNode> misses boxing needed for recursion

3. Given two sorted linked lists where one list is empty and the other is 0 -> 2 -> 4 -> null, what is the merged list? How does the merge function handle this edge case?

let l1: Option = None;
let l2 = Some(Box::new(ListNode { val: 0, next: Some(Box::new(ListNode { val: 2, next: Some(Box::new(ListNode { val: 4, next: None })) })) }));

let merged = merge_two_lists(l1, l2);

easy

A. 0 -> 2 -> 4 -> null

B. null

C. 0 -> null

D. 2 -> 4 -> null

Solution

1. Step 1: Check input lists

   l1 is None (empty), l2 is 0 -> 2 -> 4 -> null.

2. Step 2: Merge function returns non-empty list

   When one list is None, function returns the other list as is.

3. Final Answer:

   0 -> 2 -> 4 -> null -> Option A

4. Quick Check:

   Empty + list = list unchanged [OK]

Hint: If one list empty, merged list is the other list [OK]

Common Mistakes:

• Returning null when one list is empty
• Trying to merge empty list nodes incorrectly
• Losing nodes from non-empty list

4. What is the output linked list after merging these two sorted lists? List1: 1 -> 3 -> 5 -> null List2: 2 -> 4 -> 6 -> null

fn merge_two_lists(l1: Option, l2: Option) -> Option {
    match (l1, l2) {
        (None, None) => None,
        (Some(node), None) | (None, Some(node)) => Some(node),
        (Some(mut n1), Some(mut n2)) => {
            if n1.val < n2.val {
                n1.next = merge_two_lists(n1.next.take(), Some(n2));
                Some(n1)
            } else {
                n2.next = merge_two_lists(Some(n1), n2.next.take());
                Some(n2)
            }
        }
    }
}

medium

A. 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null

B. 1 -> 3 -> 5 -> 2 -> 4 -> 6 -> null

C. 2 -> 4 -> 6 -> 1 -> 3 -> 5 -> null

D. 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> null

Solution

1. Step 1: Trace first comparison

   Compare 1 and 2: 1 < 2, so 1 is first node.

2. Step 2: Recursively merge remaining lists

   Next merge 3->5 and 2->4->6. 2 < 3, so 2 next, then 3, 4, 5, 6.

3. Final Answer:

   1 -> 2 -> 3 -> 4 -> 5 -> 6 -> null -> Option A

4. Quick Check:

   Sorted merge = 1,2,3,4,5,6 [OK]

Hint: Pick smaller head each step to merge sorted lists [OK]

Common Mistakes:

• Appending lists instead of merging sorted
• Not updating next pointers correctly
• Mixing order of nodes in merged list

5. Identify the bug in this Rust merge function and select the fix:

fn merge_two_lists(l1: Option, l2: Option) -> Option {
    match (l1, l2) {
        (None, None) => None,
        (Some(node), None) => Some(node),
        (None, Some(node)) => Some(node),
        (Some(mut n1), Some(mut n2)) => {
            if n1.val <= n2.val {
                n1.next = merge_two_lists(n1.next, Some(n2));
                Some(n1)
            } else {
                n2.next = merge_two_lists(Some(n1), n2.next);
                Some(n2)
            }
        }
    }
}

medium

A. Remove match and use if-else only

B. Change if condition to n1.val < n2.val

C. Replace Some(mut n1) with &mut n1

D. Use n1.next.take() and n2.next.take() to avoid borrowing errors

Solution

1. Step 1: Identify ownership issue

   n1.next and n2.next are Options; assigning without take() causes borrowing errors.

2. Step 2: Fix by using take()

   Using n1.next.take() and n2.next.take() moves ownership out safely for recursion.

3. Final Answer:

   Use n1.next.take() and n2.next.take() to avoid borrowing errors -> Option D

4. Quick Check:

   Use take() to move Option ownership [OK]

Hint: Use .take() to move Option values safely in Rust [OK]

Common Mistakes:

• Not using take() causes borrow checker errors
• Changing comparison operator unnecessarily
• Using references incorrectly in pattern matching