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Length of Linked List

Problem Statement:​

  • Example:

✅ Solution: Iterative Traversal​

int getLength(SinglyLinkedListNode* head) {
    int length = 0;
    SinglyLinkedListNode* current = head;

    // Traverse the list and count nodes
    while(current != nullptr) {
        length++;
        current = current->next;
    }

    return length;
}

📝 How It Works​

  • We start with a pointer current at the head.
  • For every node we visit, we increment length by 1.
  • When current becomes nullptr (end of list), we stop and return the count.

This is a simple linear traversal from head to tail.

🧩 Key Logic​

There’s no recurrence or DP here — just this iterative loop:

while(current != nullptr) {
    length++;
    current = current->next;
}

⏱️ Time & Space Complexity​

MetricValue
🕒 TimeO(N), where N is the number of nodes
🧠 SpaceO(1), constant space

⚠️ Edge Cases​

  • Empty list (head == nullptr) → Length is 0
  • Single node → Length is 1

💡 Other Approaches​

ApproachTimeSpaceNotes
Iterative ✅O(N)O(1)Most efficient and preferred
RecursiveO(N)O(N)Adds call stack space (not ideal)

🔁 Related Problems​

  • Detect Length of Cycle in Linked List (Floyd’s Cycle Detection)
  • Remove N-th Node from End (need length first)
  • Check if Length is Even or Odd
  • Reverse a Linked List
💬

Discussion & Doubts