Notebook

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.

Source

Code

In [1]:

def has_cycle(head):
    '''Using set
        Time Complexity: O(n)
        Space Complexity O(n)
    '''
    if head is None:
        return False
    buff_set = {head}
    current_node = head
    while current_node.next:
        current_node = current_node.next
        if current_node in buff_set:
            return True
        buff_set.add(current_node)
    return False

Follow up:

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

Code

In [2]:

def has_cycle(head):
    '''Using 2 pointers
     space complexity can be reduced to O(1) by considering 2 pointers at different speed
     - a slow pointer (moves 1 step at a time )
     - a fast pointer (moves 2 steps at a time)

     Cycle + case A: fast pointer is 1 step behind at time T
                     at T+1, they meet up
     Cycle + Case B: fast pointer is 2 steps behind at time T
                     at T+1 they are in case A situation
                     at T+2 they meet up

    Time Complexity: O(N+K) = O(n) with:
            N = non-cyclic length
            K = cyclic length
    Space Complexity O(1)
    '''
    if head is None:
        return False
    slow = head
    fast = head.next
    while fast:
        if fast.next is None:
            return False
        fast = fast.next.next
        slow = slow.next
        if slow == fast:
            return True
    return False