Linked List Cycle

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.

Examples

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.

Constraints:

• The number of the nodes in the list is in the range $[0, 10^4].$
• $-105 <= Node.val <= 105$
• $pos is -1 or a valid index in the linked-list.$

Algorithm

1. Initialization:

   • Initialize two pointers, slow and fast, both pointing to the head of the linked list.

2. Traversal:

   • Move the slow pointer one step at a time.
   • Move the fast pointer two steps at a time.

3. Cycle Detection:

   • If there is a cycle, the fast pointer will eventually meet the slow pointer within the cycle.
   • If there is no cycle, the fast pointer will reach the end of the list (NULL).

4. Return Result:

   • If the fast pointer meets the slow pointer, return true indicating a cycle is detected.
   • If the fast pointer reaches the end of the list, return false indicating no cycle is detected.

Code Implementation

C++

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

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

        return false;
    }
};

Python

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def hasCycle(self, head: ListNode) -> bool:
        fast = head
        slow = head

        while fast is not None and fast.next is not None:
            fast = fast.next.next
            slow = slow.next
            if fast == slow:
                return True

        return False

Java

public class Solution {
    public boolean hasCycle(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;

        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) {
                return true;
            }
        }

        return false;
    }
}

JavaScript

var hasCycle = function (head) {
  let fast = head;
  let slow = head;

  while (fast !== null && fast.next !== null) {
    fast = fast.next.next;
    slow = slow.next;
    if (fast === slow) {
      return true;
    }
  }

  return false;
};