Tortoise and Hare Algorithm

141. Linked List Cycle

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.

Constraints:

• The number of the nodes in the list is in the range [0, 104].
• -105 <= Node.val <= 105
• pos is -1 or a valid index in the linked-list.

Follow up: Can you solve it using O(1) (i.e. constant) memory?

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Understanding the Problem

The problem, titled “Linked List Cycle” and classified as easy, is a common question in data structures and algorithms. It asks to determine if a given linked list has a cycle in it. A cycle in a linked list means there is some node in the list that can be reached again by continuously following the next pointer.

Matching the Problem Group to a Solution Group or Pattern

This problem falls under the category of linked list problems and specifically deals with the detection of cycles within a linked list. The solution pattern for this problem is often associated with Floyd’s Cycle-Finding Algorithm, also known as the “tortoise and the hare” algorithm.

Plan of Action to Solve the Problem Using Floyd’s Cycle Finding Algorithm

The plan to solve this problem involves using two pointers, one slow (the tortoise) and one fast (the hare). The slow pointer moves one step at a time while the fast pointer moves two steps at a time. If there is a cycle in the list, the fast pointer will eventually meet the slow pointer again. If there is no cycle, the fast pointer will reach the end of the list.

Implement Code in Java

Here is a simple implementation of the solution in Java:

public class Solution {
    public boolean hasCycle(ListNode head) {
        if (head == null || head.next == null) {
            return false;
        }
        ListNode slow = head;
        ListNode fast = head.next;
        while (slow != fast) {
            if (fast == null || fast.next == null) {
                return false;
            }
            slow = slow.next;
            fast = fast.next.next;
        }
        return true;
    }
}

Review Implementation

In the code above, we first check if the list is empty or has only one node. If so, we return false as there can’t be a cycle. We then initialize our slow and fast pointers. We enter a loop where we move our slow pointer one step and our fast pointer two steps at a time. If at any point, the fast pointer meets the slow pointer, we return true indicating a cycle. If the fast pointer reaches the end of the list, we return false indicating no cycle.

Evaluate Results

The algorithm works as expected and solves the problem efficiently. It runs in O(n) time complexity, where n is the number of nodes in the list, and uses O(1) space, thus answering the follow-up question as well.