Detecting Linked List Cycle. (LeetCode)

Task description

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.

Solution

Approach 1: Using a set

We can use a set to store the nodes we have already visited. Then, we can traverse the linked list and check if the current node is in the set. If it is, we return True. If we reach the end of the linked list, we return False.

def has_cycle(head: ListNode) -> bool:
    visited = set()
    current = head
    while current:
        if current in visited:
            return True
        visited.add(current)
        current = current.next
    return False

Time complexity - O(n) Space complexity - O(n)

Approach 2: Fast and Slow Pointers

We can use two pointers to solve this problem. The fast pointer moves two steps at a time while the slow pointer moves one step at a time. If there is no cycle in the linked list, the fast pointer will reach the end of the linked list first. If there is a cycle, the fast pointer will eventually meet the slow pointer.

def has_cycle(head: ListNode) -> bool:
    slow = head
    fast = head
    while fast and fast.next:
        slow = slow.next
        fast = fast.next.next
        if slow == fast:
            return True
    return False

Time complexity - O(n) Space complexity - O(1)