Find Middle Element of Linked List in DSA C++

Mental Model

To find the middle, move two pointers at different speeds so when the fast one reaches the end, the slow one is in the middle.

Analogy: Imagine two friends walking on a path: one walks twice as fast as the other. When the faster friend reaches the end, the slower friend is halfway there.

head -> 1 -> 2 -> 3 -> 4 -> 5 -> null
slow ↑
fast ↑

Dry Run Walkthrough

Input: list: 1 -> 2 -> 3 -> 4 -> 5

Goal: Find the middle element of the linked list

Step 1: Initialize slow and fast pointers at head (node 1)

head -> [slow->1, fast->1] -> 2 -> 3 -> 4 -> 5 -> null

Why: Start both pointers at the beginning to begin traversal

Step 2: Move slow by one node, fast by two nodes

head -> 1 -> [slow->2] -> 3 -> [fast->3] -> 4 -> 5 -> null

Why: Fast moves twice as fast to reach the end quicker

Step 3: Move slow by one node, fast by two nodes

head -> 1 -> 2 -> [slow->3] -> 4 -> 5 -> [fast->5] -> null

Why: Continue moving pointers until fast reaches the end

Step 4: Fast cannot move two nodes ahead (end reached), stop

head -> 1 -> 2 -> [slow->3] -> 4 -> 5 -> null

Why: Slow pointer is now at the middle node

Result:

1 -> 2 -> [3] -> 4 -> 5 -> null
Middle element is 3

Annotated Code

DSA C++

#include <iostream>
using namespace std;

struct Node {
    int data;
    Node* next;
    Node(int val) : data(val), next(nullptr) {}
};

Node* findMiddle(Node* head) {
    if (!head) return nullptr; // empty list
    Node* slow = head;
    Node* fast = head;
    while (fast && fast->next) {
        slow = slow->next; // move slow by one
        fast = fast->next->next; // move fast by two
    }
    return slow; // slow is at middle
}

int main() {
    // Create linked list 1->2->3->4->5->null
    Node* head = new Node(1);
    head->next = new Node(2);
    head->next->next = new Node(3);
    head->next->next->next = new Node(4);
    head->next->next->next->next = new Node(5);

    Node* middle = findMiddle(head);
    if (middle) cout << middle->data << endl;
    else cout << "List is empty" << endl;
    return 0;
}

if (!head) return nullptr; // empty list

handle empty list edge case

while (fast && fast->next) {

loop until fast reaches end or one before end

slow = slow->next; // move slow by one

advance slow pointer by one node

fast = fast->next->next; // move fast by two

advance fast pointer by two nodes

return slow; // slow is at middle

slow pointer now points to middle node

Output

3

Complexity Analysis

Time: O(n) because we traverse the list once with two pointers

Space: O(1) because only two pointers are used regardless of list size

vs Alternative: Naive approach counts nodes first then traverses again to middle, costing O(2n); this method finds middle in one pass saving time

Edge Cases

empty list

returns nullptr and prints 'List is empty'

DSA C++

if (!head) return nullptr; // empty list

single element list

returns the single node as middle

DSA C++

while (fast && fast->next) {

even number of nodes

returns the second middle node (e.g., for 4 nodes returns node 3)

DSA C++

while (fast && fast->next) {

When to Use This Pattern

When you need to find the middle of a linked list efficiently, use two pointers moving at different speeds to find it in one pass.

Common Mistakes

Mistake: Moving both pointers by one step each time, resulting in traversing the whole list twice.
Fix: Move one pointer by one step and the other by two steps to find middle in one pass.

Summary

Finds the middle node of a linked list using two pointers.

Use when you want the middle element without counting nodes first.

Moving one pointer twice as fast as the other lets you find middle in one pass.

Practice

1. What is the main idea behind finding the middle element of a singly linked list using the two-pointer approach?

easy

A. Use two pointers, one moving twice as fast as the other; when the fast pointer reaches the end, the slow pointer is at the middle.

B. Count the total nodes first, then traverse again to the middle node.

C. Reverse the linked list and pick the first element.

D. Use a stack to store nodes and pop half of them.

Solution

1. Step 1: Understand the two-pointer technique

   The fast pointer moves two nodes at a time, and the slow pointer moves one node at a time.

2. Step 2: Identify the middle when fast pointer reaches the end

   When the fast pointer reaches the end (null), the slow pointer will be at the middle node.

3. Final Answer:

   Use two pointers, one moving twice as fast as the other; when the fast pointer reaches the end, the slow pointer is at the middle. -> Option A

4. Quick Check:

   Two-pointer approach = Fast and slow pointers [OK]

Hint: Move one pointer twice as fast; slow pointer lands in middle [OK]

Common Mistakes:

• Trying to reverse the list unnecessarily
• Counting nodes twice instead of one pass
• Using extra space like stack without need

2. Which of the following C++ code snippets correctly initializes the slow and fast pointers to find the middle of a linked list named head?

easy

A. ListNode* slow = head->next; ListNode* fast = head;

B. ListNode* slow = nullptr; ListNode* fast = head;

C. ListNode* slow = head; ListNode* fast = nullptr;

D. ListNode* slow = head; ListNode* fast = head;

Solution

1. Step 1: Initialize both pointers at the head

   Both slow and fast pointers should start at the head to traverse the list correctly.

2. Step 2: Avoid null initialization for slow or fast

   Initializing slow or fast to nullptr will cause errors or skip nodes.

3. Final Answer:

   ListNode* slow = head; ListNode* fast = head; -> Option D

4. Quick Check:

   Both pointers start at head [OK]

Hint: Start both pointers at head node [OK]

Common Mistakes:

• Initializing pointers to nullptr causing runtime errors
• Starting slow pointer at head->next skipping first node
• Starting fast pointer at nullptr causing immediate stop

3. What will be the output of the following C++ code snippet that finds the middle element of a linked list?

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

Node* findMiddle(Node* head) {
  Node* slow = head;
  Node* fast = head;
  while (fast != nullptr && fast->next != nullptr) {
    slow = slow->next;
    fast = fast->next->next;
  }
  return slow;
}

int main() {
  Node n1 = {1, nullptr};
  Node n2 = {2, nullptr};
  Node n3 = {3, nullptr};
  Node n4 = {4, nullptr};
  Node n5 = {5, nullptr};
  n1.next = &n2;
  n2.next = &n3;
  n3.next = &n4;
  n4.next = &n5;
  Node* mid = findMiddle(&n1);
  std::cout << mid->data << std::endl;
  return 0;
}

medium

A. 3

B. 2

C. 4

D. 5

Solution

1. Step 1: Trace slow and fast pointers through the list

   Initially, slow=1, fast=1; after first loop, slow=2, fast=3; second loop, slow=3, fast=5; next loop fast->next is nullptr, stop.

2. Step 2: Identify slow pointer position at loop end

   Slow pointer is at node with data 3, which is the middle element.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Middle element = 3 [OK]

Hint: Fast moves two steps, slow moves one; slow ends at middle [OK]

Common Mistakes:

• Assuming middle is 2 or 4 instead of 3
• Not checking fast->next before moving fast
• Confusing zero-based indexing with node values

4. Identify the error in the following code snippet that attempts to find the middle element of a linked list and fix it:

Node* findMiddle(Node* head) {
  Node* slow = head;
  Node* fast = head;
  while (fast != nullptr) {
    slow = slow->next;
    fast = fast->next->next;
  }
  return slow;
}

medium

A. Return fast pointer instead of slow.

B. The while condition should check both fast and fast->next to avoid null pointer dereference.

C. Move slow pointer two steps instead of one.

D. Initialize slow to nullptr instead of head.

Solution

1. Step 1: Analyze the while loop condition

   The loop only checks if fast != nullptr, but fast->next might be nullptr causing a crash when accessing fast->next->next.

2. Step 2: Correct the condition to check fast->next as well

   Change the loop condition to while (fast != nullptr && fast->next != nullptr) to safely move fast pointer two steps.

3. Final Answer:

   The while condition should check both fast and fast->next to avoid null pointer dereference. -> Option B

4. Quick Check:

   Check fast and fast->next before moving fast [OK]

Hint: Always check fast and fast->next in loop condition [OK]

Common Mistakes:

• Not checking fast->next causing runtime errors
• Moving slow pointer two steps incorrectly
• Returning wrong pointer at the end

5. Given a singly linked list with an even number of nodes, how can you modify the two-pointer approach to return the first middle element instead of the second middle element?

hard

A. Initialize slow pointer to head->next instead of head.

B. Move fast pointer one step at a time instead of two.

C. Initialize fast pointer to head->next instead of head.

D. Use a stack to store nodes and pop half.

Solution

1. Step 1: Understand default behavior for even length lists

   Normally, with both pointers starting at head, slow ends at the second middle node for even length lists.

2. Step 2: Modify fast pointer start to head->next

   Starting fast at head->next causes slow to stop at the first middle node when fast reaches the end.

3. Final Answer:

   Initialize fast pointer to head->next instead of head. -> Option C

4. Quick Check:

   Fast starts at head->next for first middle [OK]

Hint: Start fast at second node to get first middle in even list [OK]

Common Mistakes:

• Changing slow pointer start instead of fast
• Moving fast pointer one step breaks logic
• Using extra space unnecessarily