Get Length of Linked List in DSA C

Mental Model

Count how many nodes are in the linked list by moving from the start to the end one by one.

Analogy: Like counting people standing in a line by walking from the first person to the last and counting each one.

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

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> null

Goal: Find how many nodes are in the list

Step 1: start at head node with value 1

head -> [1 ↑] -> [2] -> [3] -> null

Why: We begin counting from the first node

Step 2: move to next node with value 2, count = 2

head -> [1] -> [2 ↑] -> [3] -> null

Why: Count each node as we move forward

Step 3: move to next node with value 3, count = 3

head -> [1] -> [2] -> [3 ↑] -> null

Why: Continue counting nodes

Step 4: next node is null, stop counting

head -> [1] -> [2] -> [3] -> null ↑

Why: Reached end of list, total count is final

Result:

head -> [1] -> [2] -> [3] -> null
Length = 3

Annotated Code

DSA C

#include <stdio.h>
#include <stdlib.h>

// Define node structure
struct Node {
    int data;
    struct Node* next;
};

// Function to create a new node
struct Node* createNode(int data) {
    struct Node* newNode = (struct Node*)malloc(sizeof(struct Node));
    newNode->data = data;
    newNode->next = NULL;
    return newNode;
}

// Function to get length of linked list
int getLength(struct Node* head) {
    int count = 0;
    struct Node* current = head;
    while (current != NULL) {
        count++;
        current = current->next;
    }
    return count;
}

int main() {
    // Create linked list 1 -> 2 -> 3 -> null
    struct Node* head = createNode(1);
    head->next = createNode(2);
    head->next->next = createNode(3);

    int length = getLength(head);
    printf("Length = %d\n", length);
    return 0;
}

while (current != NULL) {

loop through each node until end of list

count++;

increment count for each node visited

current = current->next;

advance to next node to continue counting

Output

Length = 3

Complexity Analysis

Time: O(n) because we visit each of the n nodes once to count them

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

vs Alternative: This is optimal; no simpler way than visiting each node once to count

Edge Cases

empty list (head is NULL)

function returns 0 because there are no nodes

DSA C

while (current != NULL) {

single node list

function returns 1 counting the single node

DSA C

while (current != NULL) {

When to Use This Pattern

When you need to find how many items are in a linked list, use a simple traversal counting each node until you reach the end.

Common Mistakes

Mistake: Forgetting to move to the next node inside the loop, causing an infinite loop
Fix: Add 'current = current->next;' inside the while loop to advance the pointer

Mistake: Initializing count incorrectly or outside the function
Fix: Initialize count to zero inside the function before counting

Summary

Counts the number of nodes in a linked list by traversing it.

Use when you need to know the size of a linked list.

The key is to move through each node one by one and count them.

Practice

1. What does the function to get the length of a linked list typically do?

easy

A. Reverses the linked list

B. Deletes all nodes in the linked list

C. Finds the maximum value stored in the linked list

D. Counts the number of nodes by traversing from head to end

Solution

1. Step 1: Understand the linked list traversal

   The function starts at the head node and moves through each node until it reaches the end (null).

2. Step 2: Count each node visited

   Each time the function moves to the next node, it increases a counter by one.

3. Final Answer:

   Counts the number of nodes by traversing from head to end -> Option D

4. Quick Check:

   Length = number of nodes visited [OK]

Hint: Count nodes while moving from head to null [OK]

Common Mistakes:

• Confusing length with maximum value
• Trying to reverse the list instead
• Not traversing all nodes

2. Which of the following is the correct C syntax to declare a pointer to the head of a linked list node named head?

easy

A. struct Node* head;

B. struct Node &head;

C. Node head*;

D. struct Node head;

Solution

1. Step 1: Recall pointer declaration in C

   To declare a pointer to a struct Node, use struct Node* variableName;.

2. Step 2: Match with options

   struct Node* head; correctly declares head as a pointer to struct Node.

3. Final Answer:

   struct Node* head; -> Option A

4. Quick Check:

   Pointer syntax uses * after type [OK]

Hint: Pointer declaration uses * after type name [OK]

Common Mistakes:

• Forgetting the * for pointer
• Using & which is reference in C++ not C
• Incorrect order of * and variable name

3. What will be the output of the following C code snippet?

struct Node {
    int data;
    struct Node* next;
};

int getLength(struct Node* head) {
    int count = 0;
    while (head != NULL) {
        count++;
        head = head->next;
    }
    return count;
}

int main() {
    struct Node n3 = {30, NULL};
    struct Node n2 = {20, &n3};
    struct Node n1 = {10, &n2};
    printf("%d", getLength(&n1));
    return 0;
}

medium

A. 3

B. 2

C. 1

D. 0

Solution

1. Step 1: Analyze linked list nodes

   There are three nodes: n1 -> n2 -> n3 -> NULL.

2. Step 2: Count nodes in getLength function

   The function counts each node until it reaches NULL, so count = 3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   3 nodes counted from head to end [OK]

Hint: Count nodes until next is NULL [OK]

Common Mistakes:

• Counting only head node
• Off-by-one error in counting
• Confusing pointer addresses with count

4. Find the error in this C function to get the length of a linked list:

int getLength(struct Node* head) {
    int count = 0;
    while (head->next != NULL) {
        count++;
        head = head->next;
    }
    return count;
}

medium

A. It causes a segmentation fault immediately

B. It misses counting the last node

C. It counts nodes twice

D. It returns the data of the last node

Solution

1. Step 1: Understand the loop condition

   The loop runs while head->next != NULL, so it stops before the last node.

2. Step 2: Check counting logic

   Because it stops early, the last node is not counted, so count is one less than actual length.

3. Final Answer:

   It misses counting the last node -> Option B

4. Quick Check:

   Loop condition excludes last node [OK]

Hint: Use while(head != NULL) to count all nodes [OK]

Common Mistakes:

• Using head->next in loop condition
• Not handling empty list (NULL head)
• Incrementing count after loop ends

5. You have a linked list where some nodes may have next pointers forming a cycle (loop). How would you modify the length function to avoid infinite counting?

hard

A. Stop counting after 10 nodes to avoid infinite loop

B. Count nodes normally without any change

C. Use two pointers moving at different speeds to detect a cycle before counting

D. Set all next pointers to NULL before counting

Solution

1. Step 1: Detect cycle using two pointers

   Use Floyd's cycle detection: one pointer moves one step, another moves two steps to find a loop.

2. Step 2: If cycle detected, handle counting carefully

   Once a cycle is found, count nodes up to the cycle start to avoid infinite counting.

3. Final Answer:

   Use two pointers moving at different speeds to detect a cycle before counting -> Option C

4. Quick Check:

   Cycle detection prevents infinite loops [OK]

Hint: Detect cycle first with slow and fast pointers [OK]

Common Mistakes:

• Ignoring cycle causes infinite loop
• Stopping count arbitrarily without detection
• Modifying list structure destructively