LC 141 - Linked List Cycle

Question

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?

Links

Question here and solution here

Solution

concept

We can use a hashset to track visited node.

Alternatively, we can use a two pointers (slow and fast) and see if they can meet. Note that slow will advance one step while fast will advance two step. This will make sure that the fast pointer will catch up the slow pointer if there is a loop. Since on average the fast pointer moves 1 step w.r.t. the slow pointer, it will guaranteed to catch up the slow pointer in the loop and the time complexity is $O(n)$

code

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
	"""
	hashset
	"""
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        visited = set()
        cur = head

        while cur is not None:
            if cur in visited:
                return True
            visited.add(cur)
            cur = cur.next

        return False

class Solution:
	"""
	Floyd's tortoise and Hare
	fast and slow pointer
	"""
    def hasCycle(self, head: Optional[ListNode]) -> bool:
        slow, fast = head, head

        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if slow == fast:
                return True
        
        return False

Complexity

time: $O(n)$
space: $O(n)$ if hashset, $O(1)$ if Floyd’s tortoise and hare.