# 141. Linked List Cycle — LeetCode(Python)

I got you!

# Problem:

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?

# Solution:

`if not head or not head.next:`

return False

slow, fast = head, head

while fast and fast.next:

slow = slow.next

fast = fast.next.next

if slow == fast:

return True

return False

## Explanation —

This is a very trivial problem.

- We use a fast and slow pointer approach here.
- The fast pointer travels twice as fast as the slow pointer.
- We return the node at which the two pointers meet.
- If the while loop terminates and there is no cycle, we return False.

## Time and Space Complexity —

Since, we are iterating over all elements in the array a constant number of times, we can say that our algorithm runs in linear time.

Also, we require a constant amount of extra space as we have only used a couple of pointers.

Thus, if *n* is the length of the linked list,

Time Complexity: O(n)

Space Complexity: O(1)