Search for a Value in Linked List in DSA Rust

Mental Model

We look at each item in the list one by one until we find the value or reach the end.

Analogy: Imagine flipping through pages of a book to find a specific word. You check each page in order until you find it or finish the book.

head -> 1 -> 2 -> 3 -> null
          ↑

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3, search for value 2

Goal: Find if the value 2 exists in the linked list

Step 1: start at head node with value 1

head -> [curr->1] -> 2 -> 3 -> null

Why: We begin searching from the first node

Step 2: check if current node value (1) equals 2

head -> [curr->1] -> 2 -> 3 -> null

Why: We compare current node's value to the target

Step 3: move to next node with value 2

head -> 1 -> [curr->2] -> 3 -> null

Why: Current node is not the target, so move forward

Step 4: check if current node value (2) equals 2

head -> 1 -> [curr->2] -> 3 -> null

Why: Compare current node's value to the target

Step 5: found the value 2, stop search

head -> 1 -> [curr->2] -> 3 -> null

Why: We found the target value, no need to continue

Result:

head -> 1 -> [found->2] -> 3 -> null
Result: true

Annotated Code

DSA Rust

use std::rc::Rc;
use std::cell::RefCell;

#[derive(Debug)]
struct Node {
    value: i32,
    next: Option<Rc<RefCell<Node>>>,
}

impl Node {
    fn new(value: i32) -> Rc<RefCell<Node>> {
        Rc::new(RefCell::new(Node { value, next: None }))
    }
}

fn search(head: Option<Rc<RefCell<Node>>>, target: i32) -> bool {
    let mut current = head;
    // traverse nodes until end or found
    while let Some(node_rc) = current {
        let node = node_rc.borrow();
        if node.value == target {
            return true; // found target
        }
        current = node.next.clone(); // move to next node
    }
    false // target not found
}

fn main() {
    // create linked list 1 -> 2 -> 3 -> null
    let node1 = Node::new(1);
    let node2 = Node::new(2);
    let node3 = Node::new(3);
    node1.borrow_mut().next = Some(node2.clone());
    node2.borrow_mut().next = Some(node3.clone());

    let found = search(Some(node1), 2);
    println!("{}", found);
}

while let Some(node_rc) = current {

continue traversing nodes until we reach the end

let node = node_rc.borrow();

access current node's value and next pointer

if node.value == target {

check if current node holds the target value

return true; // found target

stop search immediately when target is found

current = node.next.clone(); // move to next node

advance current pointer to next node

Output

true

Complexity Analysis

Time: O(n) because in the worst case we check each node once until we find the value or reach the end

Space: O(1) because we only use a few variables regardless of list size

vs Alternative: Compared to arrays, linked list search is similar in time but does not allow direct access, so we must traverse nodes one by one

Edge Cases

empty list

search returns false immediately because there are no nodes

DSA Rust

while let Some(node_rc) = current {

value not in list

traverses entire list and returns false at end

DSA Rust

while let Some(node_rc) = current {

value is in the first node

returns true immediately without traversing further

DSA Rust

if node.value == target {

When to Use This Pattern

When you need to find if a value exists in a linked list, use a simple traversal search pattern by checking each node one by one.

Common Mistakes

Mistake: Not moving to the next node inside the loop, causing an infinite loop
Fix: Advance the current pointer to the next node at the end of each iteration

Mistake: Checking for value after moving current pointer, skipping the first node
Fix: Check the current node's value before moving to the next node

Summary

Searches the linked list nodes one by one to find a target value.

Use when you want to check if a value exists in a linked list.

The key is to start at the head and move forward until you find the value or reach the end.

Practice

1. What is the main purpose of searching for a value in a linked list?

easy

A. To find if a specific value exists in the list

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

C. To delete the first node of the list

D. To reverse the order of nodes in the list

Solution

1. Step 1: Understand the search operation

   Searching in a linked list means checking each node to see if it contains the value we want.

2. Step 2: Identify the goal of searching

   The goal is to find whether the value exists anywhere in the list, not to modify the list.

3. Final Answer:

   To find if a specific value exists in the list -> Option A

4. Quick Check:

   Search means find value [OK]

Hint: Search means check each node for the value [OK]

Common Mistakes:

• Confusing search with insertion or deletion
• Thinking search changes the list
• Assuming search reverses the list

2. Which Rust code snippet correctly starts a search for a value in a linked list node called head?

easy

A. let current = head.as_mut();

B. let current = head.unwrap();

C. let mut current = head.unwrap();

D. let mut current = head.as_ref();

Solution

1. Step 1: Understand linked list node type in Rust

   Linked list nodes are often wrapped in Option and Box, so head is Option.

2. Step 2: Choose safe way to start traversal

   Using head.as_ref() gives an Option reference to the node without moving ownership, allowing safe traversal.

3. Final Answer:

   let mut current = head.as_ref(); -> Option D

4. Quick Check:

   Use as_ref() to borrow node safely [OK]

Hint: Use as_ref() to get a reference without moving ownership [OK]

Common Mistakes:

• Using unwrap() which can panic if list is empty
• Using as_mut() when only reading values
• Not using mut when needing to traverse

3. What will be the output of this Rust code searching for value 5 in the linked list 1 -> 3 -> 5 -> 7?

struct Node {
    value: i32,
    next: Option,
}

fn search(head: &Option, target: i32) -> bool {
    let mut current = head.as_ref();
    while let Some(node) = current {
        if node.value == target {
            return true;
        }
        current = node.next.as_ref();
    }
    false
}

fn main() {
    let list = Some(Box::new(Node { value: 1, next: Some(Box::new(Node { value: 3, next: Some(Box::new(Node { value: 5, next: Some(Box::new(Node { value: 7, next: None })) })) })) }));
    println!("{}", search(&list, 5));
}

medium

A. false

B. true

C. Compilation error

D. Runtime panic

Solution

1. Step 1: Trace the search function

   The function starts at head (value 1), checks 1 != 5, moves to 3, checks 3 != 5, moves to 5, finds 5 == 5, returns true.

2. Step 2: Confirm output of main

   Main prints the boolean returned by search, which is true.

3. Final Answer:

   true -> Option B

4. Quick Check:

   Value found in list = true [OK]

Hint: Follow nodes until value matches or list ends [OK]

Common Mistakes:

• Assuming search returns false without checking all nodes
• Confusing Option types causing compile errors
• Forgetting to use as_ref() when traversing

4. Identify the bug in this Rust function that searches for a value in a linked list:

fn search(head: &Option, target: i32) -> bool {
    let mut current = head.unwrap();
    while current.is_some() {
        if current.value == target {
            return true;
        }
        current = current.next.unwrap();
    }
    false
}

medium

A. Incorrect comparison operator used

B. Missing mut keyword for current variable

C. Using unwrap() on Option without checking can cause panic

D. Function does not return a boolean

Solution

1. Step 1: Analyze unwrap() usage

   Calling unwrap() on Option without checking if it is Some can cause the program to panic if the list is empty or next is None.

2. Step 2: Explain safe alternative

   Using as_ref() and pattern matching with while let Some(node) = current avoids panics by safely handling None cases.

3. Final Answer:

   Using unwrap() on Option without checking can cause panic -> Option C

4. Quick Check:

   Unsafe unwrap causes panic [OK]

Hint: Never unwrap Option without checking Some first [OK]

Common Mistakes:

• Ignoring possibility of None causing panic
• Not using while let for safe traversal
• Assuming unwrap always safe

5. Given a linked list where each node contains an integer, how would you modify the search function to return the position (0-based index) of the first occurrence of a target value, or -1 if not found? Choose the correct Rust code snippet.

struct Node {
    value: i32,
    next: Option,
}

fn search_position(head: &Option, target: i32) -> i32 {
    // Fill in the code
}

hard

A. let mut current = head.as_ref(); let mut index = 0; while let Some(node) = current { if node.value == target { return index; } current = node.next.as_ref(); index += 1; } -1

B. let mut current = head.unwrap(); let mut index = 0; while current.is_some() { if current.value == target { return index; } current = current.next.unwrap(); index += 1; } -1

C. let mut current = head.as_ref(); let mut index = 1; while let Some(node) = current { if node.value == target { return index; } current = node.next.as_ref(); index += 1; } 0

D. let mut current = head.as_ref(); let mut index = 0; while let Some(node) = current { if node.value != target { return index; } current = node.next.as_ref(); index += 1; } -1

Solution

1. Step 1: Initialize traversal and index

   Start with current as a reference to head and index at 0 to count positions from start.

2. Step 2: Traverse nodes checking values

   Use while let Some(node) to safely traverse. If node.value equals target, return current index. Otherwise, move to next node and increment index.

3. Step 3: Return -1 if not found

   If loop ends without finding target, return -1 to indicate absence.

4. Final Answer:

   let mut current = head.as_ref(); let mut index = 0; while let Some(node) = current { if node.value == target { return index; } current = node.next.as_ref(); index += 1; } -1 -> Option A

5. Quick Check:

   Return index if found, else -1 [OK]

Hint: Return index on match, else -1 after full traversal [OK]

Common Mistakes:

• Starting index at 1 instead of 0
• Returning index when value does not match
• Using unwrap() causing panic on empty list