Detect Cycle in a Linked List

Given a singly linked list of integers, determine whether it contains a cycle. A cycle exists if any node in the list can be traversed more than once, forming a loop.

Input:

The first line of each test case contains the elements of the singly linked list, separated by a single space, and terminated by -1. Hence, -1 would never be a list element.

The second line contains an integer 'pos', which represents the 0-indexed position in the linked list where the tail connects. If 'pos' is -1, then there is no cycle in the linked list.

Output:

Output 'true' if the linked list contains a cycle, otherwise 'false'.

Example:

Input: 3 2 0 -4 -1 1 Output: true

Input: 1 2 3 4 -1 -1 Output: false

Constraints:

0

Question

Detect Cycle in a Linked List

Given a singly linked list of integers, determine whether it contains a cycle. A cycle exists if any node in the list can be traversed more than once, forming a loop.

Input:

The first line of each test case contains the elements of the singly linked list, separated by a single space, and terminated by -1. Hence, -1 would never be a list element.

The second line contains an integer 'pos', which represents the 0-indexed position in the linked list where the tail connects. If 'pos' is -1, then there is no cycle in the linked list.

Output:

Output 'true' if the linked list contains a cycle, otherwise 'false'.

Example:

Input:
 3 2 0 -4 -1
 1
Output:
 true

Input:
 1 2 3 4 -1
 -1
Output:
 false

Constraints:

0 <= N <= 10⁶
-1 <= pos < N
-10⁹ <= data <= 10⁹ and data ≠ -1
Time Limit: 1 second

Note:

Attempt to solve this problem with a time complexity of O(N) and space complexity of O(1).

Accepted Answer

Detect cycle in a singly linked list by using Floyd's Tortoise and Hare algorithm.

Use two pointers, slow and fast, to traverse the linked list.
If there is a cycle, the fast pointer will eventually meet the slow pointer.
Time complexity: O(N), Space complexity: O(1).