Detect a Loop in Linked List

Problem Statement:

Given head, the head of a linked list, determine if the linked list has a cycle in it.

There is a cycle in a linked list if there is some node in the list that can be reached again by continuously following the next pointer. Internally, pos is used to denote the index of the node that tail's next pointer is connected to. Note that pos is not passed as a parameter.

Return true if there is a cycle in the linked list. Otherwise, return false.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 1st node (0-indexed).

Example 2:

Input: head = [1,2], pos = 0
Output: true
Explanation: There is a cycle in the linked list, where the tail connects to the 0th node.

Example 3:

Input: head = [1], pos = -1
Output: false
Explanation: There is no cycle in the linked list.

• Example:

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✅ Solution: Two Pointer / Floyd's Cycle Detection (Tortoise and Hare)

class Solution {
public:
    bool hasCycle(ListNode *head) {
        ListNode *slow = head;
        ListNode *fast = head;

        // Move slow by 1 and fast by 2 steps
        while(fast != NULL && fast->next != NULL){
            fast = fast->next->next;
            slow = slow->next;

            // If they meet, a cycle exists
            if(fast == slow) return true;
        }

        return false; // No cycle
    }
};

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📝 How It Works

• We use two pointers:
  • slow moves one step at a time.
  • fast moves two steps at a time.
• If there’s no cycle, fast will reach the end (NULL) and loop exits.
• If there is a cycle, fast and slow will eventually meet inside the cycle.

This is known as Floyd’s Cycle Detection Algorithm (Tortoise and Hare).

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🧩 Key Logic

while(fast != NULL && fast->next != NULL){
    fast = fast->next->next;
    slow = slow->next;
    if(fast == slow) return true;
}

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⏱️ Time & Space Complexity

Metric	Value
Time	O(N)
Space	O(1)

• Each pointer travels at most N steps in the worst case.
• No extra memory is used.

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⚠️ Edge Cases

• Empty list (head == NULL)
• Single node without cycle
• Self-loop (single node pointing to itself)
• Two-node cycle

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💡 Other Approaches

Method	Time	Space	Comment
HashSet Tracking	O(N)	O(N)	Store visited nodes
Floyd’s Algorithm	O(N)	O(1)	Optimal and commonly preferred

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🔁 Related Problems

• 142. Linked List Cycle II (Find the node where cycle begins)
• 876. Middle of the Linked List
• 21. Merge Two Sorted Lists

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